@@adb012 Oh, don't get me wrong. This is indeed a very cool solution. A little bit of trig and some algebra and, bam, you get the exact answer. I just have a personal vendetta against Calc. I use any other method to solve a problem out if sheer spite. 😉
When you realize that one form of calculus is the essence of testing a series of better approximations until, for all rights and purposes, the approximation is the actual value.
I had asked a question on Physics Stack Exchange, "Why does the optimal angle depend on velocity?" Where someone answered (then suddenly deleted) and told me to check out this video. And this is a realllllllllllllllllllyyyyyy coooooooollllllll video, really nice solution. Thank you!
Hmmm, you're the second commenter today who's mentioned Physics Stack Exchange. I should check it out. Strange that the reply was deleted. I wonder if there are bots that delete posts with TH-cam links in them.
@@MathyJaphy yes, they might have a bot like that I think (and for good reason lol). I posted the same comment twice just one without the link coz I thought this was going to happen. again, awesome video!
@@MathyJaphy they delete it because SE has a wonderful policy. (No sarcasm, it's good for real). This is because just posting a link adds nothing as an answer, either it should be written in comments, or if written in an answer, the link should be complimentary and not the answer itself.
The angle must be constant with respect to the ratio of the throw velocity and the height of the building (because you can scale up a trajectory into another by scaling height and velocity up but keeping gravity constant), therefore if you believe the angle isnt constant for all starting conditions (which is intuitively clear), it cant be constant with respect to building height alone
I've tried to answer this question at least 4 times in my life, with a focus on the intuitively obvious but interesting fact that the launch angle approaches 0° as the height increases. I think I managed that small problem, but frankly it wasn't very rigourous. This is grand!
@@giusepesm yes, but the formula for final velocity has no solution (square root of a negative) if the height is too negative (meaning even throwing straight up wont reach ground level)
In India, under CBSE board, the Physics NCERT books had (and perhaps still do) what are called star-problems. They are unusually hard problems not meant for everyone but only meant for students who are pretty confident about their physics and math skills (in this context - classical mechanics and differential calculus skills). I ran into this problem for the first time few decades ago and ended up solving it the hard way. Happy to see that there exists a simple and elegant solution for this seemingly easy but actually a pretty hard problem !!
@Michael Bishop 6:25 with the height and your throwing speed known (assuming its constant as said in video), you can first calculate the landing speed and with both speeds then the throw angle for max distance. So only step that is missing is to combine the 2 formulas if you want to have it in one step!
That’s the office where I finally figured out the answer during a boring meeting in the conference room. I don’t remember talking about it, but I was so pleased with myself I suppose I must have. I’m impressed that you remember.
@@MathyJaphy Sir I got the expression to find the maximum range that is, arccos {√(2gh+v²)/√(2gh+2v²)} where, h is the height of the building v is the initial velocity
Good stuff MJ. Never thought to have pondered further beyond the "45degrees is the best angle to kick/throw a ball". Luckily I came across this video because not only did you answer a question with great intuition, but you also, and more importantly, thought to have asked a terrific question. Great Job!
A leap, yes, but hardly one of faith. :-) It's a perfectly valid observation, and therefore acceptable in a proof. I expect a professor would encourage exploring the reasons for the connection, which I omitted from the video (since I don't fully understand it).
this is actually super useful for calculating maximum trajectories when developing a video game, since most video games ignore air resistance for projectiles. Technically you can just simulate the whole projectiles path, but this is more efficient and would take way less code, awesome!
Actually, you wouldn't need to conduct a series of simulations even if this elegant solution did not exist -- this problem can also be solved numerically by applying a root-finding technique to the equation presented at 1:09.
@@bocchi2403 You can also use build in physics engines if you use game engine like Unreal Engine, Unity, etc... Usually people do it that way if they use game engine, but sometimes its better to do yourself, if you dont need to do very complicated stuff and need to do something unusual
Wow! For a long time I’ve been trying to find an intuitive way to figure out the 45 deg optimal angle without the equations of motion... and now I’ve stumbled upon a general method for all heights! Thanks a lot!
One thing missing from the video's narrative is that our little stick figure is throwing as hard as it can. It wouldn't be a very interesting problem if it could "just throw harder", as so many commenters have suggested. :-)
For those who still have trouble understanding the solution, you can think of is this way: You need horizontal speed to drive the ball forward and airtime, so it has time to travel. By aiming higher up you're essentially trading your horizontal speed for airtime. If you're throwing from a roof of a building you already have x amount of guaranteed airtime due to the height, so you need to trade less of the forward speed. Therefore (with constant throwing speed) the higher you're throwing from the closer the angle will be to 0°. On the other hand, at 0 m height you have 0 guaranteed airtime so you have to trade the throwing speed equally, which you can achieve by aiming at 45°. This is due to the simple fact that at 45° the horizontal and vertical components of your throwing speed vector will be equal (sin 45° * v0 = cos 45° * v0).
I’m glad the TH-cam algorithm showed me this video, since I’ve been trying to find an elegant way to solve this problem to explain to my sister without needing calculus. Excellent video!
Beautiful. Analytic geometry has solved a lot of things in front of my eyes that looked like they were going to take a lot more work using algebra or calculus.
This is really really cool, but a few times you've treated the length of a vector to be the same as the vector itself. That obfuscated the fact that we still don't have an actual formula that can be solved exactly by plugging in the variables at the end here. Would be cool to investigate this a bit more.
What amazes me much more is that you go from the roof to your apartment, only to be in both places at once a short time later. Wow, how did you manage that? Awesome!!!
Haha. He's just imagining himself on the roof. The real guy is solid black, and the imaginary guy is faded to grey. I hoped that would clarify that there's no disconnect in the time-space continuum.
Wow, I tried to find the optimum angle of throw for a javelin thrower considering his height and average velocity of the throw when it is released from hand. And I was struggling to find the optimum angle and gave up. This happened at the time of Olympics, and now youtube decides to recommend this video. Anyway I'm pleasantly surprised.
This is wonderful. the final velocity being perpendicular to initial velocity occurs in many similar problems. We had done this in class using vectors which gives the same conclusion ( u can find the method in 200 more puzzling problems) but the area of a triangle is a great way to look at things and can be understood by those who might not have studied by vectors yet.
This problem can be re-imagined as maximizing the parabola, which happens when the hypotenuse of your triangle passes through the parabola's focal point. This then necessitates that the starting and ending points are 90 degrees offset.
I derived this formula in 1990 when playing a WWII submarine combat simulator. Firing angle to target to hit perpendicularl amidships was invtan t° based on current bearing to target if my velocity was zero. Projectile and target velocities were constants that "fell out" of the equAtion if i, the firing platform, was stationary. It made the game trivial. Thank you for reinforcing my assumptions from 30 years ago in this video. Same problem effectively. Different visualization. It made me vizualize the basis of the analog targeting computers made during the war.
Okay you got me with the guy sneaking off at the beginning. I've already learned this. Even though my haven't. But I found myself stuck watching lol. Good video :)
I had the idea of using air time and playing with velocity vectors but did not push it far. You kept at it for 1 year. I'm in awe. This is absolutely brilliant, well done!
Awesome video! I haven’t interacted with calc or physics in the past few years but this was a great way to stretch out that part of my brain again, thank you!
great video , keep going, it was fun, i remember doing calculus when asked by prof. and he just left the problem saying maths is too tough. And today i finally have a clean and intutive way to look at it
I'm more the practical guy. I would do it with a garden hose and an angle gauge. Turn the on water and change the angle till the water hits the ground at the farest point. Then I'd measure the angle of the hose at it's end. No calculus needed. In the worst case, someone gets wet. 🗣️"SORRY, THAT'S JUST PHYSICAL EXPERIMENT!"
@@elijahdschultz If I want to know the distance, yes. But it's about to find the optimal angle. And that works even if I scale it all down. Doesn't it?
@@elijahdschultz You could also use the knowledge gathered from the water's course to approximate how a different projectile of a different speed would fly with no math necessary.
wonderful animations and wonderful explanation, i love the fact that you mention that this problem assumes negligible curvature of earth xD but that does make me wonder, what if...
That's the physics answer... Assuming it's a spherical cow in a vacuum... The Engineering answer is to exploit the magnus effect to generate lift from the ball in the direction you want the ball to travel, thereby "throwing" it further.
Yes! This video was for me. Thank you! In jr high school my PE teacher told me to use a 50deg launch angle when shot putting and while that didn't sound right there are anatomical considerations to make and he wasn't my math teacher so I didn't think much of it at the time. The launch angle question stuck in my mind though, but not so much that I took the time to dig in to it. It pleases me that it has such an elegant geometrical solution. :)
Wow. Engineer and Physics teacher here. That solution is SO creative, beautiful and simple (once you know it). The real amazing stuff here is not the solution itself, but.... HOW ON EARTH DID YOU COME UP WITH IT????? This is almost a proof that P ≠ NP. This is the first video that I watch from you and before even going to your channel to see what's else in there, I liked and subscribed.
Thank you for your kind words. I was stumped by the derivative equation, so I started working out related formulas on a calculator and plugging numbers in for fun. When you do that long enough you sometimes notice patterns, and I just got lucky. I worked out the formula for the final velocity's angle, and when I plugged in some numbers, I noticed that the maximum distance occurred when the velocity angle changed by 90 degrees. With that as a conjecture, I figured out the final formula. Then it took me a long, long time to figure out how to prove that conjecture. It seems so simple in retrospect.
I actually never thought of this problem further than the case of "same level" optimal distance . Very nice! Shame you dont publish more often (yes, I know its hard)
Thanks! I'm honored by the comparison to 3b1b. The graph for this is a mess. I made the video with a javascript application that I wrote to control all the changes, so it didn't have to be very well organized. Nevertheless, here is a link to the graph: www.desmos.com/calculator/97tdosccov If you care to check out my three other videos, you'll see they all have a link to their more reasonable Desmos graphs in the description.
What I find interesting about this is that animals (including us, though I'm sure we're getting worse) make these calculations in an unfathomable parallel fashion, and once they've learned the responses of their muscular system and visual ranging seem to get it very accurately. It's plausible there's a time-like differential electrochemical relationship in the nervous system that models/reflects it. Kind of nonsense I dream up at the top of buildings
That would be interesting to study deeper. Animals have this sort of instinct, but how, why, when? We exist today because this has been a useful genetic advantage at some point in the development of the species, and as you say it's getting worse, probably because we don't need it for survival anymore, though we like using it in many sports
@@Orakwan it would be fascinating. There might be a clue to some aspect of how and when mammals in particular seem to by a mysterious mechanism "know" how to move at birth; it's generally believed humans are born in a sense too early, at an underdeveloped point, in order that our skull volume doesn't make birth mechanically impossible or harmful. Horses etc. seem able to walk almost immediately. Presuming actual knowledge isn't inherited, some kind of system must be, it could have implications for the way robots are designed. Even the most advanced ones are incapable of walking in the way living creatures do, with that level of dexterity.
This is the approach I've been looking for to solve my age old question of "what is the ideal angle to jump off of a swing to maximize horizontal distance". Time to bust out the ti-83 and get crunching some numbers, because velocity leaving a swing isn't constant based on angle, but this should get me closer than my past brute force methods.
This is some great deduction to a nice question! It still leaves me wondering though: What is the optimal angle? I have no use for arctan of V0 over Vf. I don't know my V0. It would have been nice to see a diagram for the angle against the height for some standard V0 for example to get some intuition.
So here's my take on this: The angle is calculated as arctan(V0/Vf). V0/Vf=1/√(1+2gH/V0²) . Now we need a relationship between V0² and 2gH (without actually knowing V0). Stand near the tall building and throw the stone vertically upwards as hard as you can. Estimate how close the stone gets to the top of the building. Lets take the height as xH (as a multiple of building height). From energy conservation V0²=2gxH V0²/2gH = x Throw the stone from the building at the same velocity (since we cant be sure if the velocity is going to be same i would recommend throwing it as hard as you can both times) at the angle arctan[√(x/1+x)] to get the maximum range. This is totally impractical but just wanted to share something lol.
@@Goku_is_my_idol I worked it out similarly - except I have tan(theta) = root(x/(1+x). I actually did it in terms of the reciprocal, K = 1/x. Then tan(theta) = 1/root(1+k).
1. Determine height of building: Drop a stone and measure the time it takes to land. h = 1/2gt² = 4.9m/s² * t² 2. Throw a stone as hard as you can at 0° and measure the distance on the ground. If you don't want to go downstairs for the measurement, use a protractor to look at the landing spot, measure the angle and use d = tan(θ) * h. 3. Calculate V₀: t = sqrt(2h/g) (or just use t from 1.) V₀ = d/t Now you can plug in your V₀ into the arctan equation from the video and get your optimal angle.
This was an interesting view on an age old problem. I've always considered approximately 37.5 degrees to be optimal when including air resistance, not too far off from you (whereas you didn't factor in any air resistance).
How did you get a specific optimal angle of 37.5 degrees? What angle would it be if you ignore air resistance? Does your answer not depend on the initial velocity Vo and the height H?
This method is so good! I solved this problem with my graphical calculator some time ago, but this method is way simpler. The thing that surprises me is that both ways are very different, but they both lead to the same answer! Could you make a video too about maximizing range when throwing from a hill/slope?
Thanks for the kind words. Glad you enjoyed it. Maximizing the distance when throwing from a hill is a much simpler problem to solve in the usual way, with calculus, so I'd need some sort of hook to make an interesting video. Perhaps that hook is to solve it using this same velocity triangle method, even though that may be the harder way to do it. It is quite interesting to note that the maximum distance still comes when the final velocity vector forms a right angle with the initial velocity vector. Is this true no matter what function represents the terrain? If so, that could make an interesting video...
I did not understand, but still was fun watching. This is like watching Lee Sedol vs AlphaGo when you dont even know more than the basic rules of the game.
Is this a different answer from what you would get by imagining the path as part of the path of a 45-degree throw from ground level? Can't you just figure the 45-degree path of a throw from ground level that goes through your current position and use the angle from that partial path?
Nope. Firstly, a launch from a negative x-coordinate on the ground at a 45-degree angle would maximize the distance from that starting point, but it would not maximize the distance from the real starting point, even if the path goes through the real starting point at the same velocity. Further, any imagined launch from the ground at a 45-degree angle would hit the ground at the same 45-degree angle. But the path of maximum range from the real starting point does not. There's probably a better way to explain it. Maybe someone else reading this comment could give it a try.
@@MathyJaphy are you sure? The path the stone takes from the top of the building to the river must be a part of a parabola. If we say the building is at (a,b), and the river is (n,0) and you're trying to maximize n-a distance, where both (a,b) and (n,0) are on a parabola, I can't see how this ISNT the same parabola which passes through (0,0) and has n maximal. Granted the velocity thrown at (0,0)(called V1) is not the velocity at (a,b) (called V2). But there's only a single parabola that has V2 at (a,b) and n-a maximal. This is the exact same parabola that has V1 at (0,0) and -V1 at (n,0) with n maximal. E.g. the parabola of 45 degree tangent at (0,0) AND -45 at (n,0). Problem, of course, is that we don't immediately know what (a) is. We know (b) and we know (n-a) and we know V2. I'd have to do more math, but I can't see how the angle when the stone hits the river is NOT 45 degrees when maximizing the distance from the building for a given thrown velocity. That is, we're trying to solve the problem of "if I can throw a stone at X velocity from a building Y high, how far away can the river be and I still hit it?" The answer will be the angle thrown for the max distance.. e.g M degrees to get Q feet. We're NOT solving the problem of "the river is X distance away and I'm Y high on a building, what velocity do I have to throw at to make it?" This question doesn't have a single answer, because it will be a whole bunch of angle + velocity pairs.
@@erikanybody4298 Think about the extreme case where the initial speed (at the top of the building) is small compared to the height. There’s no launch angle that could result in a 45 degree landing angle, yet we can still ask which one maximizes the distance.
Fun fact: our physics teacher put a very similar question in a exam from last year. It was hard because if you didn't come up with the exact method of maximizing the trajectory, you were on a dead end. Great video!
But, how do you know the final velocity? Yes it is gravity*time, but how do you know the time component? Since the time component is dependent to the maximum height, among other things, which is dependent on the throwing angle. Edit: Nevermind! re-watched 3:45
Ah yes, I remember doing this during my Freshman year in HS. This one question amounted to 0.5/10 points. No, it was not easy; but the policy is that the harder the question the fewer points it provided, because hard questions just serves as a cut to separate better student from the rest.
can't you just run a simulation with a range of launch angles, plugging in your building height, see which one goes farthest and then throwing at that angle?
Sure thing but i think the problem here is that you might not be carrying a PC and have that time to write/run scripts every time you wanna throw rocks. The idea would use some simple solution that can be solved at max with your phone calculator. But I guess you can also write the script to work as a phone app too so yeah...
Very elegant proof. Reminds me though of a similar problem I worked out once but begs for a more geometric explanation. Instead of maximizing distance for a particular throw velocity it was minimizing throw velocity to hit a given target point. Through some messy trig algebra you can find that the angle is exactly halfway between the line to the target and vertical, but there has to be a far more obvious and geometric way to prove that.
Yes, it's really the same problem, isn't it. The maximum distance reachable with a given initial velocity and the minimum initial velocity needed to reach a given distance are the same things. And the required launch angle is the same whichever way you look at it. You're right, it is halfway between the line to the target and vertical! I hadn't realized that. I agree there must be a nice geometric explanation. Now I'm going to have to go figure it out (unless you figure it out first)! Either way, this could be the subject for a follow-up video!
What I like the most is that when you realized brute force (aka calculus) was a dead end, you didn't give up, and struggled for one year using your imagination to come up with something more clever. Congratulations, I really liked it.
Теоретик: я бросил камень недостаточно сильно, а значит - необходимо потратить много часов на высчитывание вектора скорости и идеального угла для броска! Практик: Ладно, просто попробую снова.
I remember doing a project about this in college, in a numerical methods class. It was just a "use computer to approximate best solution" kind of project. I simply approximated the value of theta that minimizes the huge derivative shown at the start of the video. I also remember just plugging in hella values of theta and keeping the solution that went the farthest lmao this solution is so much more elegant than my crappy python code
As i am just that person standing frequently on tall buildings with a stone in my hand not knowing what brought me there or what to do with the object, this video explains very well my purpose in life.
I hate it when I am at a roof of a building that is near a river, somehow got a small random object with me and when I decide to throw it into the river, it just falls short. Thank you for the solution to this problem
Im a mechanical engineering student and i thought about exactly this+ calculated the ideal angles half a year ago lmao + if you take air resistance into count youre always getting a slightly more horizontal throw
Well, the original "Paris Gun" also considered that there is less air resistance at higher altitudes and that a big shell can actually climb above much of the atmosphere. That worked out that the "best" angle for range was higher than 45 degrees. The idea was to get the shell into space ASAP and then the horizonal velocity vector would not be as diminished by the air resistance.
Btw the solution for that should be a sin function that hits 0 at its lowest point and twice the cars speed at its highest i didnt do calculations but it made sense to me like that
This is amazing I've thought about this problem during physics what would be the optimal angle in a throw because it's always been given by the problem. And this video answered my question, physics is cool.
I found throwing off a mountain that throwing upward slightly with a flat rock i get the furthest distance. This is a 500 ft or so mountain in WV. Throwing at a 45 only caused the air to slow it's velocity to near 0 and then it fell straight down before reaching the base. But the flat rock sailed over to the other mountain crossing the whole valley below.
There's something beautiful about the way that simple intuition so often shortcuts the complexities of mathematics. Some people, like myself (and military gunners) have a pretty accurate built-in intuition for projectile trajectories. Maths, although great in many theoretical situations, and some practical solutions is, a lot of the time, a total wank. Instead of studying such shite inside the schoolroom, my mates and I were instead up on the library roof actually throwing rocks at passing cars. We soon got very good at knowing the best angle and force to throw. Meanwhile, the propeller-head kids were inside with their geometry and calculus. And I'm sure if they had ever challenged us to a contest, we'd have won every time.
I had the mathematics for ages during my elementary, high school and university years and I only wish someone had the guts to present it in an interesting way such as this. 20 years after graduating I can't remember what sin, cos ang tang means any more.
The way that I was taught was that for the maximum horizontal distance, the landing angle should be 45°, since any steeper of an angle would mean that you're going down more than you're going horizontally, and the whole goal is to go as far horizontally as possible. So if you start at the end with a 45° angle and work backward knowing your energy of the throw (initial kinetic energy plus gravitational potential energy at throw), you just trace the curve back to get the angle
That happens to be wrong, though, as this video shows. The optimal landing angle is 90° different from the optimal launch angle, neither of which is 45° unless the landing and launch spots are at the same height.
"Whatever you can solve with calculus, I can approximate with algebra."
mere geometry and sketches in CAD can also solve otherwise complex calculus problems too.
The cool stuff here is that this in not an approximation, but an exact solution.
@@adb012 Oh, don't get me wrong. This is indeed a very cool solution. A little bit of trig and some algebra and, bam, you get the exact answer.
I just have a personal vendetta against Calc. I use any other method to solve a problem out if sheer spite. 😉
When you realize that one form of calculus is the essence of testing a series of better approximations until, for all rights and purposes, the approximation is the actual value.
Newton’s method goes brrr
I had asked a question on Physics Stack Exchange, "Why does the optimal angle depend on velocity?" Where someone answered (then suddenly deleted) and told me to check out this video. And this is a realllllllllllllllllllyyyyyy coooooooollllllll video, really nice solution. Thank you!
Hmmm, you're the second commenter today who's mentioned Physics Stack Exchange. I should check it out. Strange that the reply was deleted. I wonder if there are bots that delete posts with TH-cam links in them.
@@MathyJaphy yes, they might have a bot like that I think (and for good reason lol). I posted the same comment twice just one without the link coz I thought this was going to happen. again, awesome video!
@@MathyJaphy they delete it because SE has a wonderful policy. (No sarcasm, it's good for real). This is because just posting a link adds nothing as an answer, either it should be written in comments, or if written in an answer, the link should be complimentary and not the answer itself.
Bro you in 11th?
The angle must be constant with respect to the ratio of the throw velocity and the height of the building (because you can scale up a trajectory into another by scaling height and velocity up but keeping gravity constant), therefore if you believe the angle isnt constant for all starting conditions (which is intuitively clear), it cant be constant with respect to building height alone
I've tried to answer this question at least 4 times in my life, with a focus on the intuitively obvious but interesting fact that the launch angle approaches 0° as the height increases. I think I managed that small problem, but frankly it wasn't very rigourous. This is grand!
You and me the same. Once we are sufficiently high up, the flat launch becomes best.
You have? This has been around since maths itself.
What I like about the end result is that it still works for the H=0 case (when you're on the ground), and you still get 45 degrees! Great video!
I mean if that wasn't going to work for H=0, the formula would have been proven wrong..
Does it work for negative values? (When you're below ground level that rock will land on - on a ditch for example)
Yes, it does. Around age 15-16 I worked out iirc, 48 deg was optimum for hitting the ball over a fence 4m higher than the start point 30m away
@@giusepesm yes, but the formula for final velocity has no solution (square root of a negative) if the height is too negative (meaning even throwing straight up wont reach ground level)
I did that mental check as well. A tower with height equal to 0 is just a special case of a tower of height h.
In India, under CBSE board, the Physics NCERT books had (and perhaps still do) what are called star-problems. They are unusually hard problems not meant for everyone but only meant for students who are pretty confident about their physics and math skills (in this context - classical mechanics and differential calculus skills). I ran into this problem for the first time few decades ago and ended up solving it the hard way. Happy to see that there exists a simple and elegant solution for this seemingly easy but actually a pretty hard problem !!
Thank you for taking the time to tell me about this.
So you found a root of the derivative? How?
@Michael Bishop 6:25 with the height and your throwing speed known (assuming its constant as said in video), you can first calculate the landing speed and with both speeds then the throw angle for max distance. So only step that is missing is to combine the 2 formulas if you want to have it in one step!
@@MathyJaphy this method is actually genius. I am going to use this in a computer program.
@Vishal Mishra What's the harder way to do this problem?
I remember a certain person posing this exact question 30-some odd years ago in a work office in Bethesda. Nice analysis!
That’s the office where I finally figured out the answer during a boring meeting in the conference room. I don’t remember talking about it, but I was so pleased with myself I suppose I must have. I’m impressed that you remember.
@@MathyJaphy Sir I got the expression to find the maximum range that is,
arccos {√(2gh+v²)/√(2gh+2v²)}
where, h is the height of the building
v is the initial velocity
Good stuff MJ. Never thought to have pondered further beyond the "45degrees is the best angle to kick/throw a ball". Luckily I came across this video because not only did you answer a question with great intuition, but you also, and more importantly, thought to have asked a terrific question. Great Job!
You made quite a leap of faith, going from distance to area of triangle. No math or physics professor would let you get away with it.
A leap, yes, but hardly one of faith. :-) It's a perfectly valid observation, and therefore acceptable in a proof. I expect a professor would encourage exploring the reasons for the connection, which I omitted from the video (since I don't fully understand it).
@@MathyJaphy Christians also don't fully understand their god, yet they have FAITH in him. I rest my case.
I love how geometry has a solution to everything.
Triangles, triangles everywhere I look
Truly the apex of math
Yeah. Almost everything in math is useless without geometry
@Coup Lab Geometry is the easiest way of using imaginary numbers though? All the numbers become a grid lol
And the physics, which reduced the event to a straight line➖ where both horizontal and vertical vectors were proved to be constant.
this is actually super useful for calculating maximum trajectories when developing a video game, since most video games ignore air resistance for projectiles.
Technically you can just simulate the whole projectiles path, but this is more efficient and would take way less code, awesome!
Wait what, game programmer need to study physics? The game physics is derived from real physic coded into the game?
@@bocchi2403 pretty much, yes
@@bocchi2403 how do you expect games to look and feel realistic then?
Actually, you wouldn't need to conduct a series of simulations even if this elegant solution did not exist -- this problem can also be solved numerically by applying a root-finding technique to the equation presented at 1:09.
@@bocchi2403 You can also use build in physics engines if you use game engine like Unreal Engine, Unity, etc...
Usually people do it that way if they use game engine, but sometimes its better to do yourself, if you dont need to do very complicated stuff and need to do something unusual
What a genius solution! I would've gone brute force calculus and then get stuck trying to solve for the angle
My mind is absolutely blown. This is so elegant and wonderful. Keep up the good work!
Wow! For a long time I’ve been trying to find an intuitive way to figure out the 45 deg optimal angle without the equations of motion... and now I’ve stumbled upon a general method for all heights! Thanks a lot!
One thing missing from the video's narrative is that our little stick figure is throwing as hard as it can. It wouldn't be a very interesting problem if it could "just throw harder", as so many commenters have suggested. :-)
Yeah then take a stronger stick figure.
For those who still have trouble understanding the solution, you can think of is this way:
You need horizontal speed to drive the ball forward and airtime, so it has time to travel. By aiming higher up you're essentially trading your horizontal speed for airtime. If you're throwing from a roof of a building you already have x amount of guaranteed airtime due to the height, so you need to trade less of the forward speed. Therefore (with constant throwing speed) the higher you're throwing from the closer the angle will be to 0°. On the other hand, at 0 m height you have 0 guaranteed airtime so you have to trade the throwing speed equally, which you can achieve by aiming at 45°. This is due to the simple fact that at 45° the horizontal and vertical components of your throwing speed vector will be equal (sin 45° * v0 = cos 45° * v0).
With this formula you can calculate perfect angle to get out of hole amaizing i must say
I’m glad the TH-cam algorithm showed me this video, since I’ve been trying to find an elegant way to solve this problem to explain to my sister without needing calculus. Excellent video!
My physics teacher just finished the projectile motion chapter, and you concluded everything in it in 7 minutes
Beautiful. Analytic geometry has solved a lot of things in front of my eyes that looked like they were going to take a lot more work using algebra or calculus.
This is really really cool, but a few times you've treated the length of a vector to be the same as the vector itself. That obfuscated the fact that we still don't have an actual formula that can be solved exactly by plugging in the variables at the end here. Would be cool to investigate this a bit more.
obfuscated
@@TerraBlo Yes, obfuscated. Perspicuous as the daylight sky!
Obfuscated 2
you look absolutely ripped in a tank top bro
I like the fact as a non native speaker, that there are English words even native speakers despise xD Love it!
What amazes me much more is that you go from the roof to your apartment, only to be in both places at once a short time later. Wow, how did you manage that? Awesome!!!
Haha. He's just imagining himself on the roof. The real guy is solid black, and the imaginary guy is faded to grey. I hoped that would clarify that there's no disconnect in the time-space continuum.
It was uploaded 6 months ago, and I'm finding it now. TH-cam, why you never recommended this channel before!
A very unique solution indeed.
Incredible video. Really happy that you're continuing to make videos too, 5 months later.
Wow,
I tried to find the optimum angle of throw for a javelin thrower considering his height and average velocity of the throw when it is released from hand. And I was struggling to find the optimum angle and gave up.
This happened at the time of Olympics, and now youtube decides to recommend this video.
Anyway I'm pleasantly surprised.
This is wonderful. the final velocity being perpendicular to initial velocity occurs in many similar problems. We had done this in class using vectors which gives the same conclusion ( u can find the method in 200 more puzzling problems) but the area of a triangle is a great way to look at things and can be understood by those who might not have studied by vectors yet.
I just watched three of your videos consecutively, each more unique and nerdy than the last. This is beautiful stuff, thank you for making it!
This problem can be re-imagined as maximizing the parabola, which happens when the hypotenuse of your triangle passes through the parabola's focal point. This then necessitates that the starting and ending points are 90 degrees offset.
That's a solid idea
Back to your crayons. Haha,
As an army vet I just can’t accept that this intelligent comment came from a Marine.
@@BillBrasky368 That's the trick: the purple ones make you smarter.
I derived this formula in 1990 when playing a WWII submarine combat simulator.
Firing angle to target to hit perpendicularl amidships was invtan t° based on current bearing to target if my velocity was zero.
Projectile and target velocities were constants that "fell out" of the equAtion if i, the firing platform, was stationary.
It made the game trivial.
Thank you for reinforcing my assumptions from 30 years ago in this video.
Same problem effectively. Different visualization.
It made me vizualize the basis of the analog targeting computers made during the war.
This is more interesting than my whole year of physics class
Okay you got me with the guy sneaking off at the beginning. I've already learned this. Even though my haven't. But I found myself stuck watching lol. Good video :)
Very cool approach to this ubiquitous physics problem! And also very well presented!
I had the idea of using air time and playing with velocity vectors but did not push it far.
You kept at it for 1 year. I'm in awe.
This is absolutely brilliant, well done!
Well, a year of thinking about it now and then. It wouldn't let go of me! Glad you enjoyed it. Thanks for the kind words.
i must be always standing on the ground as this presentation is way over my head
If you're standing on the ground then be sure to throw the small projectile at a 45 degree angle 😂
Awesome video! I haven’t interacted with calc or physics in the past few years but this was a great way to stretch out that part of my brain again, thank you!
This was really good video. All the points presented are so intuitive that my mind was in awe. Thanks and got subscribed :D.
great video , keep going, it was fun, i remember doing calculus when asked by prof. and he just left the problem saying maths is too tough. And today i finally have a clean and intutive way to look at it
I'm more the practical guy. I would do it with a garden hose and an angle gauge. Turn the on water and change the angle till the water hits the ground at the farest point. Then I'd measure the angle of the hose at it's end.
No calculus needed.
In the worst case, someone gets wet.
🗣️"SORRY, THAT'S JUST PHYSICAL EXPERIMENT!"
This approach only works if you throw the ball at the same speed that water leaves the hose.
@@elijahdschultz If I want to know the distance, yes. But it's about to find the optimal angle. And that works even if I scale it all down. Doesn't it?
@@sigisalmen2399 unfortunately no. As the video demonstrates, the optimal angle depends on how fast you can throw.
@@elijahdschultz I guess now I can't avoid to get on the rooftop and make some passing stranger wet.
Have a nice day
@@elijahdschultz You could also use the knowledge gathered from the water's course to approximate how a different projectile of a different speed would fly with no math necessary.
I’m so glad this video exists. I use to completely not even understand how to throw rocks from a building at a perfect angle, and now I still don’t.
wonderful animations and wonderful explanation, i love the fact that you mention that this problem assumes negligible curvature of earth xD but that does make me wonder, what if...
Ooh! Hold my beer…
It is
if the earths curvature was modelled, the solutions wouldn't behave nicely. I mean you could throw it so fast it would orbit forever
@@prototypeinheritance515 Wow that's a cool observation. A discontinuous function arising from physics. I haven't seen such a thing before.
And the Jeopardy answer is: What is a satellite?
This video has strong 3b1b and minutephysics vibes. Way to go man!
That's the physics answer... Assuming it's a spherical cow in a vacuum...
The Engineering answer is to exploit the magnus effect to generate lift from the ball in the direction you want the ball to travel, thereby "throwing" it further.
The Computer Science answer is to write a simulator and run it until it is 99.99% sure its the perfect angle.
Biology answer?
@@truthseeker7815 adrenaline
@@catchara1496, oh yeah, Fight or Flight State
@@catchara1496 Philosophy answer?
Yes! This video was for me. Thank you! In jr high school my PE teacher told me to use a 50deg launch angle when shot putting and while that didn't sound right there are anatomical considerations to make and he wasn't my math teacher so I didn't think much of it at the time. The launch angle question stuck in my mind though, but not so much that I took the time to dig in to it. It pleases me that it has such an elegant geometrical solution. :)
Wow. Engineer and Physics teacher here. That solution is SO creative, beautiful and simple (once you know it). The real amazing stuff here is not the solution itself, but.... HOW ON EARTH DID YOU COME UP WITH IT????? This is almost a proof that P ≠ NP. This is the first video that I watch from you and before even going to your channel to see what's else in there, I liked and subscribed.
Thank you for your kind words. I was stumped by the derivative equation, so I started working out related formulas on a calculator and plugging numbers in for fun. When you do that long enough you sometimes notice patterns, and I just got lucky. I worked out the formula for the final velocity's angle, and when I plugged in some numbers, I noticed that the maximum distance occurred when the velocity angle changed by 90 degrees. With that as a conjecture, I figured out the final formula. Then it took me a long, long time to figure out how to prove that conjecture. It seems so simple in retrospect.
no teacher writes like this
@@Kiba114 What do you mean?
@@Kiba114 ?
@@Kiba114 wut?
The first minute you described, I did *EXACTLY*. Reached a dead end, even with WolframAlpha. I'm dying 😂
I actually never thought of this problem further than the case of "same level" optimal distance . Very nice! Shame you dont publish more often (yes, I know its hard)
I'm very surprised I never came across this version of the problem before. Very cool solution!!
This is great! Like 3Blue1Brown but with Desmos. Could you post as a reply the Desmos graph you used to make this?
Thanks! I'm honored by the comparison to 3b1b.
The graph for this is a mess. I made the video with a javascript application that I wrote to control all the changes, so it didn't have to be very well organized. Nevertheless, here is a link to the graph: www.desmos.com/calculator/97tdosccov
If you care to check out my three other videos, you'll see they all have a link to their more reasonable Desmos graphs in the description.
You just can't imagine how much I needed this!
What I find interesting about this is that animals (including us, though I'm sure we're getting worse) make these calculations in an unfathomable parallel fashion, and once they've learned the responses of their muscular system and visual ranging seem to get it very accurately. It's plausible there's a time-like differential electrochemical relationship in the nervous system that models/reflects it. Kind of nonsense I dream up at the top of buildings
That would be interesting to study deeper. Animals have this sort of instinct, but how, why, when? We exist today because this has been a useful genetic advantage at some point in the development of the species, and as you say it's getting worse, probably because we don't need it for survival anymore, though we like using it in many sports
@@Orakwan it would be fascinating. There might be a clue to some aspect of how and when mammals in particular seem to by a mysterious mechanism "know" how to move at birth; it's generally believed humans are born in a sense too early, at an underdeveloped point, in order that our skull volume doesn't make birth mechanically impossible or harmful. Horses etc. seem able to walk almost immediately. Presuming actual knowledge isn't inherited, some kind of system must be, it could have implications for the way robots are designed. Even the most advanced ones are incapable of walking in the way living creatures do, with that level of dexterity.
This is the approach I've been looking for to solve my age old question of "what is the ideal angle to jump off of a swing to maximize horizontal distance". Time to bust out the ti-83 and get crunching some numbers, because velocity leaving a swing isn't constant based on angle, but this should get me closer than my past brute force methods.
Dude.. you're underrated
This is epic! My first thought was to use newtons method to approximate the the angle using the derivative but this is so much simpler!
I didn’t understand a damn thing, but I like your funny words magic man.
I literally tried solving this problem with calculus with my classmate today and when I came home this video appeared!
This is some great deduction to a nice question!
It still leaves me wondering though: What is the optimal angle? I have no use for arctan of V0 over Vf. I don't know my V0. It would have been nice to see a diagram for the angle against the height for some standard V0 for example to get some intuition.
So here's my take on this:
The angle is calculated as arctan(V0/Vf).
V0/Vf=1/√(1+2gH/V0²) .
Now we need a relationship between V0² and 2gH (without actually knowing V0).
Stand near the tall building and throw the stone vertically upwards as hard as you can.
Estimate how close the stone gets to the top of the building.
Lets take the height as xH (as a multiple of building height).
From energy conservation
V0²=2gxH
V0²/2gH = x
Throw the stone from the building at the same velocity (since we cant be sure if the velocity is going to be same i would recommend throwing it as hard as you can both times) at the angle arctan[√(x/1+x)] to get the maximum range.
This is totally impractical but just wanted to share something lol.
@@Goku_is_my_idol I worked it out similarly - except I have tan(theta) = root(x/(1+x). I actually did it in terms of the reciprocal, K = 1/x. Then tan(theta) = 1/root(1+k).
@@tedsheridan8725 yes you're right
It should be arctan[√(x/1+x)]
I did it in my head while writing the comment hence the miscalculation
1. Determine height of building:
Drop a stone and measure the time it takes to land. h = 1/2gt² = 4.9m/s² * t²
2. Throw a stone as hard as you can at 0° and measure the distance on the ground. If you don't want to go downstairs for the measurement, use a protractor to look at the landing spot, measure the angle and use d = tan(θ) * h.
3. Calculate V₀:
t = sqrt(2h/g) (or just use t from 1.)
V₀ = d/t
Now you can plug in your V₀ into the arctan equation from the video and get your optimal angle.
Omg please upload some more videos like this i absolutely love it
This was an interesting view on an age old problem. I've always considered approximately 37.5 degrees to be optimal when including air resistance, not too far off from you (whereas you didn't factor in any air resistance).
How did you get a specific optimal angle of 37.5 degrees? What angle would it be if you ignore air resistance? Does your answer not depend on the initial velocity Vo and the height H?
The final velocity equation can also be determined using kinematics equations for free falling bodies (that means we ignore air restistance).
Omg thank you so much I spent whole day yesterday trying to throw a stone to the damn river till neigbours got to the roof and kicked my ass!
This method is so good! I solved this problem with my graphical calculator some time ago, but this method is way simpler. The thing that surprises me is that both ways are very different, but they both lead to the same answer! Could you make a video too about maximizing range when throwing from a hill/slope?
Thanks for the kind words. Glad you enjoyed it. Maximizing the distance when throwing from a hill is a much simpler problem to solve in the usual way, with calculus, so I'd need some sort of hook to make an interesting video. Perhaps that hook is to solve it using this same velocity triangle method, even though that may be the harder way to do it. It is quite interesting to note that the maximum distance still comes when the final velocity vector forms a right angle with the initial velocity vector. Is this true no matter what function represents the terrain? If so, that could make an interesting video...
I did not understand, but still was fun watching. This is like watching Lee Sedol vs AlphaGo when you dont even know more than the basic rules of the game.
This is some prime content. Very beautifully done.
Is this a different answer from what you would get by imagining the path as part of the path of a 45-degree throw from ground level? Can't you just figure the 45-degree path of a throw from ground level that goes through your current position and use the angle from that partial path?
Nope. Firstly, a launch from a negative x-coordinate on the ground at a 45-degree angle would maximize the distance from that starting point, but it would not maximize the distance from the real starting point, even if the path goes through the real starting point at the same velocity. Further, any imagined launch from the ground at a 45-degree angle would hit the ground at the same 45-degree angle. But the path of maximum range from the real starting point does not. There's probably a better way to explain it. Maybe someone else reading this comment could give it a try.
@@MathyJaphy , Good point. I stand corrected.
@@MathyJaphy are you sure? The path the stone takes from the top of the building to the river must be a part of a parabola.
If we say the building is at (a,b), and the river is (n,0) and you're trying to maximize n-a distance, where both (a,b) and (n,0) are on a parabola, I can't see how this ISNT the same parabola which passes through (0,0) and has n maximal.
Granted the velocity thrown at (0,0)(called V1) is not the velocity at (a,b) (called V2). But there's only a single parabola that has V2 at (a,b) and n-a maximal. This is the exact same parabola that has V1 at (0,0) and -V1 at (n,0) with n maximal. E.g. the parabola of 45 degree tangent at (0,0) AND -45 at (n,0).
Problem, of course, is that we don't immediately know what (a) is. We know (b) and we know (n-a) and we know V2.
I'd have to do more math, but I can't see how the angle when the stone hits the river is NOT 45 degrees when maximizing the distance from the building for a given thrown velocity.
That is, we're trying to solve the problem of "if I can throw a stone at X velocity from a building Y high, how far away can the river be and I still hit it?" The answer will be the angle thrown for the max distance.. e.g M degrees to get Q feet.
We're NOT solving the problem of "the river is X distance away and I'm Y high on a building, what velocity do I have to throw at to make it?" This question doesn't have a single answer, because it will be a whole bunch of angle + velocity pairs.
@@erikanybody4298 Think about the extreme case where the initial speed (at the top of the building) is small compared to the height. There’s no launch angle that could result in a 45 degree landing angle, yet we can still ask which one maximizes the distance.
Fun fact: our physics teacher put a very similar question in a exam from last year. It was hard because if you didn't come up with the exact method of maximizing the trajectory, you were on a dead end.
Great video!
But, how do you know the final velocity? Yes it is gravity*time, but how do you know the time component? Since the time component is dependent to the maximum height, among other things, which is dependent on the throwing angle.
Edit: Nevermind! re-watched 3:45
Wow! This is a great problem with a fantastic solution. Thank you for sharing. Subscribed!
Ah yes, I remember doing this during my Freshman year in HS. This one question amounted to 0.5/10 points. No, it was not easy; but the policy is that the harder the question the fewer points it provided, because hard questions just serves as a cut to separate better student from the rest.
Maximum when range when vertical displacement was not zero was on your Freshman HS school test???
Excellent presentation! And very thorough. Thanks, MJ.
can't you just run a simulation with a range of launch angles, plugging in your building height, see which one goes farthest and then throwing at that angle?
Sure thing but i think the problem here is that you might not be carrying a PC and have that time to write/run scripts every time you wanna throw rocks. The idea would use some simple solution that can be solved at max with your phone calculator. But I guess you can also write the script to work as a phone app too so yeah...
Very elegant proof. Reminds me though of a similar problem I worked out once but begs for a more geometric explanation. Instead of maximizing distance for a particular throw velocity it was minimizing throw velocity to hit a given target point. Through some messy trig algebra you can find that the angle is exactly halfway between the line to the target and vertical, but there has to be a far more obvious and geometric way to prove that.
Yes, it's really the same problem, isn't it. The maximum distance reachable with a given initial velocity and the minimum initial velocity needed to reach a given distance are the same things. And the required launch angle is the same whichever way you look at it. You're right, it is halfway between the line to the target and vertical! I hadn't realized that. I agree there must be a nice geometric explanation. Now I'm going to have to go figure it out (unless you figure it out first)! Either way, this could be the subject for a follow-up video!
If I had a pound for every time i heard "assuming ideal condition" I wouldn't have to be at university right now
The ability to idealize and approximate and still get a useful result is the most important skill of a physicist.
@@atalazs OK....?
What I like the most is that when you realized brute force (aka calculus) was a dead end, you didn't give up, and struggled for one year using your imagination to come up with something more clever. Congratulations, I really liked it.
Теоретик: я бросил камень недостаточно сильно, а значит - необходимо потратить много часов на высчитывание вектора скорости и идеального угла для броска!
Практик: Ладно, просто попробую снова.
Never been taught such an elegant solution, especially without calculus.
45°. Easy. Artillery pieces with lower angles are called cannons, pieces with higher angles mortars. And a howitzer shouts in both angle groups.
Apparently you didn't watch the video or you'd know it's not 45°.
@@thorr18BEM Of course I didn't watch the video. What sense does it make to solve a puzzle after seeing the solution?
@@Nikioko So you don’t even know the question to begin with.
Cool.
Gotta love watching a video with massive helpful knowledge in it and having understanded nothing from it.
Very woke not to have used the old "muzzle velocity" metaphor.
I swear next time I am on top of a building I will never ever think of throwing anything… this has exhausted my brain !!
:-) th-cam.com/video/qQXs27H7n1U/w-d-xo.html
MJ Thank you for your very elegant post two years ago, that I only recently came across
I remember doing a project about this in college, in a numerical methods class. It was just a "use computer to approximate best solution" kind of project. I simply approximated the value of theta that minimizes the huge derivative shown at the start of the video. I also remember just plugging in hella values of theta and keeping the solution that went the farthest lmao this solution is so much more elegant than my crappy python code
I will come back to this video when I have learned more about physics
really well done with the animations and everything, i thought the channel would have a lot more subs!!!
As i am just that person standing frequently on tall buildings with a stone in my hand not knowing what brought me there or what to do with the object, this video explains very well my purpose in life.
Such a great video! Intuitively I came up with the same answer but had no idea why.
I hate it when I am at a roof of a building that is near a river, somehow got a small random object with me and when I decide to throw it into the river, it just falls short.
Thank you for the solution to this problem
Brilliant presentation and explanation! Great work!
Im a mechanical engineering student and i thought about exactly this+ calculated the ideal angles half a year ago lmao
+ if you take air resistance into count youre always getting a slightly more horizontal throw
Well, the original "Paris Gun" also considered that there is less air resistance at higher altitudes and that a big shell can actually climb above much of the atmosphere. That worked out that the "best" angle for range was higher than 45 degrees. The idea was to get the shell into space ASAP and then the horizonal velocity vector would not be as diminished by the air resistance.
Just curious, are you saying that you arrived at the same method?
Yes
When im bored i think about the most random stuff like the horizontal velocity of a point of a tire etc idk whats wrong with me
Btw the solution for that should be a sin function that hits 0 at its lowest point and twice the cars speed at its highest i didnt do calculations but it made sense to me like that
This is amazing I've thought about this problem during physics what would be the optimal angle in a throw because it's always been given by the problem. And this video answered my question, physics is cool.
I found throwing off a mountain that throwing upward slightly with a flat rock i get the furthest distance. This is a 500 ft or so mountain in WV. Throwing at a 45 only caused the air to slow it's velocity to near 0 and then it fell straight down before reaching the base. But the flat rock sailed over to the other mountain crossing the whole valley below.
This is amazing! Keep up the good work :)
This was amazing dude, great work!
Nice! I think about this every time I water the garden. Right on
So many aha moments with this one, which is the best feeling!
Thank you for making this
There's something beautiful about the way that simple intuition so often shortcuts the complexities of mathematics. Some people, like myself (and military gunners) have a pretty accurate built-in intuition for projectile trajectories. Maths, although great in many theoretical situations, and some practical solutions is, a lot of the time, a total wank. Instead of studying such shite inside the schoolroom, my mates and I were instead up on the library roof actually throwing rocks at passing cars. We soon got very good at knowing the best angle and force to throw. Meanwhile, the propeller-head kids were inside with their geometry and calculus. And I'm sure if they had ever challenged us to a contest, we'd have won every time.
The power of triangle, nice illustration btw want some more like this video. Up.
I'm so happy I discovered this channel! Next to Mathologer and 2blue1brown, this is exactly what I enjoy. :)
I searched this and I am glad that I found an answer
gee, its so good to learn another useful trick in algebra! I'm sure I will use it as often as I use Bhaskara's formula on my daily life
I had the mathematics for ages during my elementary, high school and university years and I only wish someone had the guts to present it in an interesting way such as this. 20 years after graduating I can't remember what sin, cos ang tang means any more.
The way that I was taught was that for the maximum horizontal distance, the landing angle should be 45°, since any steeper of an angle would mean that you're going down more than you're going horizontally, and the whole goal is to go as far horizontally as possible. So if you start at the end with a 45° angle and work backward knowing your energy of the throw (initial kinetic energy plus gravitational potential energy at throw), you just trace the curve back to get the angle
That happens to be wrong, though, as this video shows. The optimal landing angle is 90° different from the optimal launch angle, neither of which is 45° unless the landing and launch spots are at the same height.
Man, I'm definitely gonna give this a try next time I'm on top of a shoreside building with a frictionless projectile.
In an alternate universe, I can understand this. That makes me happy.