Bruh this man just saved my grade... My physics teacher gave me a random distance and told me to find vi and angle needed and I had no clue what in the world I was doing...
Hi Michel. Fellow Physics teacher here. Have you considered using a different trig substitution method? If you substitute tan θ = sin θ / cos θ and use tan^2 θ = sec^2 θ - 1 to substitute 1/cos^2 θ with (tan^2 θ + 1) , then you will have a quadratic in tan θ which you can solve with the quadratic formula. This will also yield two possible solutions for the angle, both of which should be valid. I use the aforementioned method in my class and I think it is a little easier to work with than the phase angle method. I want to make sure there isn't a reason I am missing for why you do not use this method.
There are often multiple ways to solve a problem, and obviously, I didn't pick the most direct method. Sometimes one picks the first thing that comes to mind which is what I did apparently. :)
Well done Sir, well done. I was almost convinced that the solution will be trial and error; vary theta until X is the desired distance. That's beautiful.
So glad I found this. I'm using this equation to program a game. I have a question. There must be some initial velocity that is too low so that no possible angle will make the projectile reach x. But I can't see how/if the equation blows up in this situation. (There should be no solution, right?) How do we know if this is the case? I can see that initial velocity of zero will blow up the denominator, but there should be some positive low initial velocities that have no solution. Also, shouldn't there sometimes be 2 solutions? For example, firing directly downward towards the target or firing upward at a steep angle and letting gravity pull the projectile back down to the same target. How would this show up in the equation?
Hello, I know I'm late but in case this is useful to someone... Well, the cos function can only give a number between 1 and -1. So the equation won't have any solution if the whole thing inside the arccos is not in [-1, 1]. In other words, you probably want an "if" that checks that number first, and then do the arccos after you are sure that it won't break.
quite the problem! very unexpected arriving at that identity, well done. there's also an alternate from of the answer that uses the sine addition formula instead of cosine. the answers look very similar, as to be expected with trig identities
Hi, Please recommend a good Physics textbook for me (perhaps a supplement to your videos). I want to explore the following topics: a) Kinematics (including 2D motion) b) Forces c) Rotational motion d) Work, energy and power e) Fluid mechanics Thanks and keep up the good work, sir!
Hi Michel. Thanks for the interesting lessons. I met some issues with the current solution method. It works well until we have a positive "h". The projectile source is above the target, but if we have a negative "h" it doesn't work. I assume the main reason for this archcosine function.
Would you still approach this the same way if the projectile had a downward angle? (ie. If a volleyball player hit the ball at a downward angle, and you wanted to find the downward angle so that it would just make it inbound)
Hello teacher. Thank you very much for putting so much effort in your videos. It is loable to see people selflessly helping others. Although the content of the video was a bit intricate, you exposed it pretty clear. Nonetheless, I can appreciate how little i know of trigonometry. Is there any good trigonometry book you could recommend?
Thanks again. I´ll see the videos but I´ll also keep on looking for a book to consult every now and then and study basic concepts, that are not taught, in some depth.
@@nahuel3256 A great book with plenty of trigonometry is "A Graphical Approach to Pre-Calculus" (Hornsby et al) www.amazon.com/Graphical-Approach-Precalculus-Limits-Books/dp/0134698223/ref=dp_ob_title_bk
Hello teacher. How can I find the shooting angle of a ball to hit a basketball hoop, if I have the speed, gravitational acceleration and vertical and horizontal distances? I can't deduce an equation to find the angle when the target is above the launch point.
Amazing video and very useful formula! It helped me a lot, thank you! I want to use this formula in my project. Is there any book where this formula is written in? My project teacher told me only the written books' or their scan copies' references are allowed in a project. Maybe you know where I can find this formula? Could you please help me?
4:07 I was confused because I thought the h+ was part of the numerator and I was so confused how you were canceling out v naught with the h there. Might want to adjust the line height or put parenthesis next time.
Shooting a basketball into a hoop from a distance of 10 m with a relative height of 2 m and a release speed of 7 m/s Inputting these values into your equation gives me a cosine ratio larger than one! so i can't calculate the arccos. Help appreciated.
That probably means that with an intial speed of 7 m/sec you cannot reach the basket. Do a quick check with the range equation (with an angle of 45 degrees for the maximumm range, when basket and starting point are at the same height). R = Vo^2/g sin(2 theta) = (7^2/9.8) sin (90) = 49/9.8 = 5 Which means the maximum range would be 5 meters.
my case is to determine the angle of a projectil which shoots towards a target of XYZ coordinate in space. so in the expression of the angle that you have just determined, my Y will not be zero, is that it?
Thank you for the lesson teacher, but i have a couple of questions. First one is that if im lower than the target/destination can i input the height from where is shoot as a negative because it is In relation To the target. Second one is that there is two angles that work, one that goes (if start and end on level ground) More than 45 degrees and one that goes less than 45. This calculation calculates the More than 45 one and i would like To know the one that is less than 45. This goes with the third question. Because this calculates More than 45 it Will always be positive (right?) and if i wanted the one with less than 45 it would give a negative degree. Would that brake the calculation or would all of those be possible To find? Thank you for the lesson!
For your first question it is arbitrary where you put your "zero height". It will make no difference on your answer. For you second question, it is possible on some problems that there are 2 answers for the angle and in some cases there is only one answer for the angle.
Is there are good reason why not to use the tan (theta) and solve the quadratic for tan (theta) before evaluating arctan of these two solutions? That's been my normal approach and it's how I've taught it but just curious. Excellent videos here, it's a very clean explanation.
The required angle can also be deteremined by the range equation (provided the projectile is launched from and land at the same height). The range equation is: R = [Vo ^2 x sin (2 theta) ] / |g|
can you pls explain to me finding the angle on projectile at different hight the video dont make sense to me i saw you usepower reducing identity still i dont understand
It would be difficult to explain in a comment what took 13+ minutes on the video and a white board. We'll plan on making a few more videos on this topic starting with some easier examples in the future.
I struggled too to find this identity in any of my Maths books. Good video - I could never really remember these types being that difficult. I wonder if i iterated guesses earlier on to try come close to balancing the equation.
Hello, thank you for this video it helps me a lot in my research. but if you allow me, I have a question: here you took Y (final height) equal zero. if in an exercise we have the value of this Y (final height) how will the expression of the angle become? (of course taking into account the values of Yo(h), Vo, X, G...) thank you in advance sir
If the final height is not zero, then the constant in the quadratic quation would be different but the method of solving the problem would not change at all.
Thank you very much sir, but I think there is a Problem, most of the time there is more than 1 Solution for the Angle. How can i always get the smallest Angle ?
Yes, there may be 2 solutions. On a flat surface with the projectile starting at ground height, it will go the farthest at a 45 degree angle. But it will go the same distance at a 30 degree angle and at a 60 degree angle. If you get 2 solutions then both angles will give you the same distance.
The height of the landing is not important, just the difference of the height between the starting height and the finishing height. The method will be the same.
@@MichelvanBiezen I worked it out but thank you for replying anyway. I have a follow up question, The equation is now affected by drag, the way drag is calculated is as follows Velocity at time = drag coefficient * velocity every frame There are 60 frames in a second, to get the the point where the projectile hits we need to use the example below: Calculate the projectile trajectory starting at 0, 0 with an initial velocity of 70 and an angle of 45 After 1.666ms have passed (1 second/frames in a second) save that position and current angle the projectile is going and multiply it’s velocity by 0.999 (drag coefficient), recalculate the trajectory using these new values Repeat this by going another 1.666ms down and saving position and current angle, velocity is now multiplied by 0.999 again, recalculating the projectile trajectory. Repeat this until we hit the ground. First question: is how can I turn this into an equation (if possible)? Second question: how can I calculate the required elevation angle to hit the target accounting for this new projectile behaviour? Thanks for your time I have been slamming my head against a wall trying to solve it, good luck.
Thanks for your hard work teacher but I have a couple of doubts on how you came up with this final equation because I myself derived this equation on my own and I ended up with this: 1/2cos^-1[((-2h+2(gx^2/2v^2)((x^2+h^2)^1/2))/(x^2+h^2)]+tan^-1(x/h)}} long ass equation, before you ask me questions on how I came up with this equation, I observed a big mistake you made during your 8th step of the derivation, instead of -h/2(1-cos2theta) it was supposed to be[[[ -h/2(cos2theta)]]](correction!!!!) because I checked this myself on a scientific calculator then I noticed that you messed up on your double angle identitie and that's how I came up with a much different equation. Thanks alot!!!!! btw, I hope you look forward to replying to my message soon. bye!!!!!!
in this equation if we put h = 0 , can we arrive at the exact same result for angle theta when it is projected from ground ??? I desperately need an answer. please...
When h = 0 this becomes much easier because y0 would then be equal to 0, so you won't have to worry about the h being in the way during the first equation.
How would you find the initial angle with wind? So the velocityX would be affected by wind the same way as the velocityY is getting affected by gravity.
@@ericbridge8419 I have no clue how to calculate it (, wouldn't have been able to solve this one without the video either). It is more complicated than this :D. I think it is a 3-dimensional case where the wind affects to velocities X and Z depending from what angle it blows (if it blows directly from behind, it would accelerate vX and if it blows from 90 degree angle from the side it would affect vZ like gravity affects vY). I think you must know the weight of the thrown/shot object to determine how much does the wind affect it's trajectory, also. Here I make the assumption the wind blows directly from a side and not obliquely from above or below when it should also affect vY (?) It would be cool to see how exactly it's done but damn, this is already sth i wouldn't have come up with myself:D
P1ker1 My case was just 2d, with only x and y involved, no 3rd dimension and no z. You have: -gravity -particles mass -wind angle is always horizontal -how much the particle is getting affected by wind -initial speed of the particle -the point that the shot needs to hit I needed to know it because I wanted to write a hack for the tank game called "Shellshock live". A hack that would calculate the angles that I needed to shoot at to hit the target. I made a completely working one that only calculates it with no wind. I asked this question in 4 forums too and no anwsers lol. I never made my dream come true :(
@@MichelvanBiezen I figured it out. We were given the distance between the base of a cliff and the point where the projectile lands. But the time given is the time from the initial point to a point A. That point A is at the same height of the initial point. So basically, Yo and Yfa cancel out and using the equation Yf = Yo + Vyo + at²/2 I can solve for my angle since Vo is given and time A is given too. At the end I would get theta = arcsin(at²/Vo)
Hey Sir i wanted to ask that if i differentiate and set the final equation to 0 will that give me optimal angle that gives furthest distance? If not than how would i go about that
I need to be able to prove that in the same instance particle A is thrown from height on the horizontal at an angle Theta and a particle B is thrown from height below the horizontal from point hj at an angle of Beta with half the velocity of A and prove that they will never come in contact. So essentially I need to get quadratic simultaneous equations with t and Theta & Beta and show them as an inequality. anyone able to help me with this?!
my problem wants the height of a ball B with Vo=4 in the way that from this height it will meet a ball A thrown upwords with an initaial velocity 20.The distance beetweem two is 4m.And the ball B is thrown horizontally.the balls are throuwn at the same time. Help me please?🙏
You first find the time it takes for the ball thrown horizontally to reach the ball thrown upward: x = v * t (4 = 4 * t) Therefore t = 1 sec. Then you determine where each ball will be after 1 sec and set that height equal. Essentially the ball thrown horizontally will fall down in the vertical direction like a ball that is just dropped. (Now it is just like the example in one of the videos in the playlist)
I appreciate the help, but for some reason the formula does not work for me. I'm trying to solve an equation where h = 37.5m, x = 230.22m, velocity-initial = 40.53m/s, t = 5.68s and gravity works differently so that downwards acceleration = 18m/s^2. Is the equation I'm attempting to solve impossible?
@@MichelvanBiezen Essentially, yes. Fortunately, I did a bit of experimenting and realized that I solved for the initial velocity incorrectly, and the formula worked properly once I did the math right. Thank you for explaining the formula, I never would've worked it out on my own (the problem is not for any particular class, I just wanted to find the angle at which a projectile could be launched so that it traveled the farthest in a hypothetical situation, and I wasn't sure if the problem I posed for myself would be plausible)
This is terribly convoluted. To find that angle for this type of problem: [arcsin(xg/2(Vo^2))]/2. What I'm still not understanding is why the back of my book gives 2 answers, and the 2nd is not as simple as 90 degrees minus this first angle...
I want to use this formula for my video game to calculate the initial angle to throw a stone to an enemy with a const. velocity, with a certain x-distance between them and a height difference. For some unknown reason the arc-cos is higher than 1, so the result is undefined. Do you know why my numerator is greater than the denominator? Or in which case that happens? Thanks :)
Oh wow! I doubled the initial velocity and now the curve looks great! Thank you so much for given me the hint! By increasing the velocity the numerator gets greater, but do have a special meaning for this? Was the velocity too low so there cant be a solution?
I mean, if you solve this with the method our lecturer showed us, the solution would be the quadratic equation like this: (V0² +- sqrt(V0⁴-2V0²*h*g-g²*x²)) Theta = arctan ( -------------------------------------------------) ( (g*x) ) Of course, this implies that there are 2 possible solutions to this problem. So again, my question is, how do I find the second solution using your method? :)
@@larsb.1972 Hi! First of all, thank you very much for the equation, that your lecturer gave you. You can get the first solution by using + in front of the sqrt and the second by using - in front of the sqrt.
@@RUMPshit Hi. Thank you for the answer! I used that equation to show that this problem has two solutions, I know how that one works :D I was talking about the equation he derives in this video. It seems to me that Michel's formula is only solvable for one of the two solutions
hi, i need helps solving a really hard problem, what if the H = L SIN (ANGLE) while L is 23cm and the angle is the angle that we looking for in the video, please help me
im in 6th grade im trynna make a calculator for a video game i dont wanna do the algebra from here 😭😭😭😭😭😭😭😭😭 thx anyway tho im prob gonna cook with this oh nvm its kinda easy carry on
![[Watch Party] RoV Pro League 2024 Winter | with Moowan ( Bacon Time Vs Talon )](http://i.ytimg.com/vi/CCEDhu6dPN8/mqdefault.jpg)
Bruh this man just saved my grade... My physics teacher gave me a random distance and told me to find vi and angle needed and I had no clue what in the world I was doing...
This saved my day. I was struggling on how to implement a function for the flight of a projectile to a certain location this whole day. Thanks a lot!
You're welcome 😊
Thanks Michel for letting us know the precise angle so that the prisoners of the 9th isle can accurately find the angle to hit the elite mercs 👍
Interesting application. 🙂
Hi Michel. Fellow Physics teacher here. Have you considered using a different trig substitution method? If you substitute tan θ = sin θ / cos θ and use tan^2 θ = sec^2 θ - 1 to substitute 1/cos^2 θ with (tan^2 θ + 1) , then you will have a quadratic in tan θ which you can solve with the quadratic formula. This will also yield two possible solutions for the angle, both of which should be valid. I use the aforementioned method in my class and I think it is a little easier to work with than the phase angle method. I want to make sure there isn't a reason I am missing for why you do not use this method.
There are often multiple ways to solve a problem, and obviously, I didn't pick the most direct method. Sometimes one picks the first thing that comes to mind which is what I did apparently. :)
can you explain how you did this please?
your channel is a coder’s best friend THANK YOU MICHEL!!
You are welcome. Glad you like it.
this is amazing sir; I was trying to solve a similar problem for a project and this video helped a lot.
Great. Glad you found our videos! 🙂
Well done Sir, well done. I was almost convinced that the solution will be trial and error; vary theta until X is the desired distance. That's beautiful.
Trial and error is a valid method, but this will give you the answer directly 🙂
Honestly, Im struggling with physics 11, your videos are more advanced but there full of great knowledge, thank you!
bru am using his vids to get ez extra marks LOLOLOLOLOLOLOL
So glad I found this. I'm using this equation to program a game. I have a question. There must be some initial velocity that is too low so that no possible angle will make the projectile reach x. But I can't see how/if the equation blows up in this situation. (There should be no solution, right?) How do we know if this is the case? I can see that initial velocity of zero will blow up the denominator, but there should be some positive low initial velocities that have no solution.
Also, shouldn't there sometimes be 2 solutions? For example, firing directly downward towards the target or firing upward at a steep angle and letting gravity pull the projectile back down to the same target. How would this show up in the equation?
Haaha me too what game engine y use ?
Hello, I know I'm late but in case this is useful to someone... Well, the cos function can only give a number between 1 and -1. So the equation won't have any solution if the whole thing inside the arccos is not in [-1, 1].
In other words, you probably want an "if" that checks that number first, and then do the arccos after you are sure that it won't break.
quite the problem! very unexpected arriving at that identity, well done. there's also an alternate from of the answer that uses the sine addition formula instead of cosine. the answers look very similar, as to be expected with trig identities
Yes, that is actually a better (easier) way to approach the problem. Good point.
I always get the biggest clue to solve stuff from your video sir, Thank you
Wouldn't -h/2 disappear if you were to move it to the h?
I'm deeply grateful ! Your video help me solve my game balistic problem :)
Hi,
Please recommend a good Physics textbook for me (perhaps a supplement to your videos). I want to explore the following topics:
a) Kinematics (including 2D motion)
b) Forces
c) Rotational motion
d) Work, energy and power
e) Fluid mechanics
Thanks and keep up the good work, sir!
There are many good physics books on the market. If you can buy an older edition on line you will get the best bargain.
Michel van Biezen Thanks, but are there any specific books you would recommend?
Engineering dynamics by hibbeler 14th edition.
Hi Michel.
Thanks for the interesting lessons.
I met some issues with the current solution method.
It works well until we have a positive "h". The projectile source is above the target, but if we have a negative "h" it doesn't work.
I assume the main reason for this archcosine function.
It should also work if h is negative.
Thank you!! You're an amazing teacher!
Thank u .
u made me fall in love in physics
I've learned so much from you thank you so much. How I wish you were my uncle for I am so interested in the subject that you have an expertise with.
Keep that enthusiasm going. Learning science and engineering can be very exciting.
I was so confident that I could derive this until I've tried for a whole day
This is soo scary..
This is indeed a challenging problem.
Would you still approach this the same way if the projectile had a downward angle? (ie. If a volleyball player hit the ball at a downward angle, and you wanted to find the downward angle so that it would just make it inbound)
super incredible.........too innovative method of solving!!
Hello teacher. Thank you very much for putting so much effort in your videos. It is loable to see people selflessly helping others.
Although the content of the video was a bit intricate, you exposed it pretty clear. Nonetheless, I can appreciate how little i know of trigonometry. Is there any good trigonometry book you could recommend?
There are 3 good playlists on trigonometry on this channel starting with this one: TRIGONOMETRY BASICS - PRECALCULUS 6
Thanks again. I´ll see the videos but I´ll also keep on looking for a book to consult every now and then and study basic concepts, that are not taught, in some depth.
@@nahuel3256 A great book with plenty of trigonometry is "A Graphical Approach to Pre-Calculus" (Hornsby et al) www.amazon.com/Graphical-Approach-Precalculus-Limits-Books/dp/0134698223/ref=dp_ob_title_bk
Hello teacher. How can I find the shooting angle of a ball to hit a basketball hoop, if I have the speed, gravitational acceleration and vertical and horizontal distances?
I can't deduce an equation to find the angle when the target is above the launch point.
There are examples in the playlist just like that. Find the equation for the distance in both the x and y direction and solve them simultaneously.
@@MichelvanBiezen Sorry teacher. I played a game of basketball. I need to find the launch angle. But I can not.
Can you help me, please?
I would like to put your name on my graduation work.
Amazing video and very useful formula! It helped me a lot, thank you! I want to use this formula in my project. Is there any book where this formula is written in? My project teacher told me only the written books' or their scan copies' references are allowed in a project. Maybe you know where I can find this formula? Could you please help me?
You can find this example in most physics books.
4:07 I was confused because I thought the h+ was part of the numerator and I was so confused how you were canceling out v naught with the h there. Might want to adjust the line height or put parenthesis next time.
Yes, parentheses would have been useful here.
Love from India🇨🇮
Thank you very much!
I was trying to figure this out starting from the same equations and getting stuck midway.
Shooting a basketball into a hoop from a distance of 10 m with a relative height of 2 m and a release speed of 7 m/s Inputting these values into your equation gives me a cosine ratio larger than one! so i can't calculate the arccos. Help appreciated.
That probably means that with an intial speed of 7 m/sec you cannot reach the basket. Do a quick check with the range equation (with an angle of 45 degrees for the maximumm range, when basket and starting point are at the same height). R = Vo^2/g sin(2 theta) = (7^2/9.8) sin (90) = 49/9.8 = 5 Which means the maximum range would be 5 meters.
This was very interesting! Thank you!
Massive respect to you SIR!
my case is to determine the angle of a projectil which shoots towards a target of XYZ coordinate in space. so in the expression of the angle that you have just determined, my Y will not be zero, is that it?
The range will be R = sqrt( x^2 + y^2) which then becomes the "x" in this problem. This assumes that z is the point above the ground.
Thank you for the lesson teacher, but i have a couple of questions. First one is that if im lower than the target/destination can i input the height from where is shoot as a negative because it is In relation To the target. Second one is that there is two angles that work, one that goes (if start and end on level ground) More than 45 degrees and one that goes less than 45. This calculation calculates the More than 45 one and i would like To know the one that is less than 45. This goes with the third question. Because this calculates More than 45 it Will always be positive (right?) and if i wanted the one with less than 45 it would give a negative degree. Would that brake the calculation or would all of those be possible To find? Thank you for the lesson!
For your first question it is arbitrary where you put your "zero height". It will make no difference on your answer. For you second question, it is possible on some problems that there are 2 answers for the angle and in some cases there is only one answer for the angle.
You're the man!! Excellent explanation
Is there are good reason why not to use the tan (theta) and solve the quadratic for tan (theta) before evaluating arctan of these two solutions? That's been my normal approach and it's how I've taught it but just curious. Excellent videos here, it's a very clean explanation.
There are often multiple ways in which a problem can be solved.
Sir please upload Required angle solution for incline projectile motion
The required angle can also be deteremined by the range equation (provided the projectile is launched from and land at the same height). The range equation is: R = [Vo ^2 x sin (2 theta) ] / |g|
Very, VERY useful. Many thanks. I was close but those trigonometric identities, specially the second one, made it impossible for me :c
can you pls explain to me finding the angle on projectile at different hight the video dont make sense to me i saw you usepower reducing identity still i dont understand
It would be difficult to explain in a comment what took 13+ minutes on the video and a white board. We'll plan on making a few more videos on this topic starting with some easier examples in the future.
thanks sir for your this amazing guidance to Radha Singh from India
At 9:30, why (a)sin(x) + (b)cos(x) = sqrt(a^2 + b^2)cos(x - arctan(a/b))
That is a trig identity which can be found in any trig book
I struggled too to find this identity in any of my Maths books. Good video - I could never really remember these types being that difficult. I wonder if i iterated guesses earlier on to try come close to balancing the equation.
Hello, thank you for this video it helps me a lot in my research. but if you allow me, I have a question: here you took Y (final height) equal zero. if in an exercise we have the value of this Y (final height) how will the expression of the angle become?
(of course taking into account the values of Yo(h), Vo, X, G...) thank you in advance sir
If the final height is not zero, then the constant in the quadratic quation would be different but the method of solving the problem would not change at all.
Thank you very much sir, but I think there is a Problem, most of the time there is more than 1 Solution for the Angle. How can i always get the smallest Angle ?
Yes, there may be 2 solutions. On a flat surface with the projectile starting at ground height, it will go the farthest at a 45 degree angle. But it will go the same distance at a 30 degree angle and at a 60 degree angle. If you get 2 solutions then both angles will give you the same distance.
@@MichelvanBiezen thank you for the fast reply 🙂
But I'm not sure if I understand at what part of the equation I can get a second solution?
How did changing the launch height of the projectile affect its time of flight, maximum height, and range?
You work the problem out the exact same way as any other projectile problem with the only difference that the initial height is h instead of zero.
@@MichelvanBiezen Thankyou so much
Woww amazing! Thank you very much!
You're very welcome! Glad you found our videos. 🙂
What if “Y” is not equal to 0
The height of the landing is not important, just the difference of the height between the starting height and the finishing height. The method will be the same.
@@MichelvanBiezen I worked it out but thank you for replying anyway. I have a follow up question,
The equation is now affected by drag, the way drag is calculated is as follows
Velocity at time = drag coefficient * velocity every frame
There are 60 frames in a second, to get the the point where the projectile hits we need to use the example below:
Calculate the projectile trajectory starting at 0, 0 with an initial velocity of 70 and an angle of 45
After 1.666ms have passed (1 second/frames in a second) save that position and current angle the projectile is going and multiply it’s velocity by 0.999 (drag coefficient), recalculate the trajectory using these new values
Repeat this by going another 1.666ms down and saving position and current angle, velocity is now multiplied by 0.999 again, recalculating the projectile trajectory.
Repeat this until we hit the ground.
First question: is how can I turn this into an equation (if possible)?
Second question: how can I calculate the required elevation angle to hit the target accounting for this new projectile behaviour?
Thanks for your time I have been slamming my head against a wall trying to solve it, good luck.
Thanks for your hard work teacher but I have a couple of doubts on how you came up with this final equation because I myself derived this equation on my own and I ended up with this: 1/2cos^-1[((-2h+2(gx^2/2v^2)((x^2+h^2)^1/2))/(x^2+h^2)]+tan^-1(x/h)}} long ass equation, before you ask me questions on how I came up with this equation, I observed a big mistake you made during your 8th step of the derivation, instead of -h/2(1-cos2theta) it was supposed to be[[[ -h/2(cos2theta)]]](correction!!!!) because I checked this myself on a scientific calculator then I noticed that you messed up on your double angle identitie and that's how I came up with a much different equation. Thanks alot!!!!! btw, I hope you look forward to replying to my message soon. bye!!!!!!
in this equation if we put h = 0 , can we arrive at the exact same result for angle theta when it is projected from ground ??? I desperately need an answer. please...
When h = 0 this becomes much easier because y0 would then be equal to 0, so you won't have to worry about the h being in the way during the first equation.
How would you find the initial angle with wind? So the velocityX would be affected by wind the same way as the velocityY is getting affected by gravity.
So how do you calculate it?? I'm still looking for it xD
@@ericbridge8419
I have no clue how to calculate it (, wouldn't have been able to solve this one without the video either). It is more complicated than this :D.
I think it is a 3-dimensional case where the wind affects to velocities X and Z depending from what angle it blows (if it blows directly from behind, it would accelerate vX and if it blows from 90 degree angle from the side it would affect vZ like gravity affects vY). I think you must know the weight of the thrown/shot object to determine how much does the wind affect it's trajectory, also.
Here I make the assumption the wind blows directly from a side and not obliquely from above or below when it should also affect vY (?)
It would be cool to see how exactly it's done but damn, this is already sth i wouldn't have come up with myself:D
P1ker1 My case was just 2d, with only x and y involved, no 3rd dimension and no z.
You have:
-gravity
-particles mass
-wind angle is always horizontal
-how much the particle is getting affected by wind
-initial speed of the particle
-the point that the shot needs to hit
I needed to know it because I wanted to write a hack for the tank game called "Shellshock live". A hack that would calculate the angles that I needed to shoot at to hit the target. I made a completely working one that only calculates it with no wind. I asked this question in 4 forums too and no anwsers lol. I never made my dream come true :(
@@ericbridge8419 Add/subtract your wind velocity (depending on direction) to your Vx
Fantastic
Glad you liked it.
It works with negative angles? I need find an angle to shot to down
Yes, it works the same way. The initial velocity in the y direction will be negative.
@@MichelvanBiezen thanks teacher! You are a genious
How about if both initial height and end height is not equal to zero ?
calculate the difference and then draw it as one zero
Hi sir, i would like to know how to solve the equation if the y is not zero?
Exactly the same as this video.
How do you solve it if height is not given, but time is given
Work it out exactly the same way, but replace t with the time and solve for h
@@MichelvanBiezen I figured it out. We were given the distance between the base of a cliff and the point where the projectile lands. But the time given is the time from the initial point to a point A. That point A is at the same height of the initial point. So basically, Yo and Yfa cancel out and using the equation Yf = Yo + Vyo + at²/2 I can solve for my angle since Vo is given and time A is given too. At the end I would get theta = arcsin(at²/Vo)
Hey Sir i wanted to ask that if i differentiate and set the final equation to 0 will that give me optimal angle that gives furthest distance? If not than how would i go about that
Yes, that would be one way to approach it.
I need to be able to prove that in the same instance particle A is thrown from height on the horizontal at an angle Theta and a particle B is thrown from height below the horizontal from point hj at an angle of Beta with half the velocity of A and prove that they will never come in contact. So essentially I need to get quadratic simultaneous equations with t and Theta & Beta and show them as an inequality. anyone able to help me with this?!
my problem wants the height of a ball B with Vo=4 in the way that from this height it will meet a ball A thrown upwords with an initaial velocity 20.The distance beetweem two is 4m.And the ball B is thrown horizontally.the balls are throuwn at the same time.
Help me please?🙏
You first find the time it takes for the ball thrown horizontally to reach the ball thrown upward: x = v * t (4 = 4 * t) Therefore t = 1 sec. Then you determine where each ball will be after 1 sec and set that height equal. Essentially the ball thrown horizontally will fall down in the vertical direction like a ball that is just dropped. (Now it is just like the example in one of the videos in the playlist)
@@MichelvanBiezen thank youuu soo much :)you're the best!!
@@Emma-od4qx ma'am I would like to contact you. Would you like the same please??
Thank you
I appreciate the help, but for some reason the formula does not work for me. I'm trying to solve an equation where h = 37.5m, x = 230.22m, velocity-initial = 40.53m/s, t = 5.68s and gravity works differently so that downwards acceleration = 18m/s^2. Is the equation I'm attempting to solve impossible?
What is the problem asking you to solve? Find the angle at which the projectile is fired?
@@MichelvanBiezen Essentially, yes. Fortunately, I did a bit of experimenting and realized that I solved for the initial velocity incorrectly, and the formula worked properly once I did the math right. Thank you for explaining the formula, I never would've worked it out on my own (the problem is not for any particular class, I just wanted to find the angle at which a projectile could be launched so that it traveled the farthest in a hypothetical situation, and I wasn't sure if the problem I posed for myself would be plausible)
Great! We often learn through trial and error and by sticking to it.
This is terribly convoluted. To find that angle for this type of problem: [arcsin(xg/2(Vo^2))]/2. What I'm still not understanding is why the back of my book gives 2 answers, and the 2nd is not as simple as 90 degrees minus this first angle...
How much do i need to pay to make this guy my prof instead ?
I want to use this formula for my video game to calculate the initial angle to throw a stone to an enemy with a const. velocity, with a certain x-distance between them and a height difference. For some unknown reason the arc-cos is higher than 1, so the result is undefined. Do you know why my numerator is greater than the denominator? Or in which case that happens? Thanks :)
What happens if you increase your initial velocity?
Oh wow! I doubled the initial velocity and now the curve looks great! Thank you so much for given me the hint! By increasing the velocity the numerator gets greater, but do have a special meaning for this? Was the velocity too low so there cant be a solution?
That is correct.
Sir, shouldn't theta have 2 values ?? because there can be two paths for the same destination.
The second path requires a different initial velocity.
thank you sir!
You are welcome!
Hi, Michel. Doesn't this scenario have two possible solutions? How do I find the other one?
I mean, if you solve this with the method our lecturer showed us, the solution would be the quadratic equation like this:
(V0² +- sqrt(V0⁴-2V0²*h*g-g²*x²))
Theta = arctan ( -------------------------------------------------)
( (g*x) )
Of course, this implies that there are 2 possible solutions to this problem. So again, my question is, how do I find the second solution using your method? :)
@@larsb.1972 Hi! First of all, thank you very much for the equation, that your lecturer gave you. You can get the first solution by using + in front of the sqrt and the second by using - in front of the sqrt.
Here you can see the formula in action: www.desmos.com/calculator/ochebzfymt
@@RUMPshit Hi. Thank you for the answer! I used that equation to show that this problem has two solutions, I know how that one works :D I was talking about the equation he derives in this video. It seems to me that Michel's formula is only solvable for one of the two solutions
@@larsb.1972 Your equation is far better than Michael's. It actually saved me from hours of research, i couldn't find anything like it lol.
hi, i need helps solving a really hard problem, what if the H = L SIN (ANGLE) while L is 23cm and the angle is the angle that we looking for in the video, please help me
What is L? Is it the range of the projectile?
I just realized this was mechanics
I loved this video, but I didn’t understand it 🙁
What if final height isnt 0...
Then you work out the problem in exactly the same way, and you insert the final height instead of zero.
@@MichelvanBiezen so from the beginning you work with final height Y instead of 0 right?
That is correct.
There is one another case, when ball is thrown from ground and falls at a height. Where can I watch this case problem?
We have examples in the playlist where projectile has to clear a height in the distance (same thing), in the playlist.
im in 6th grade im trynna make a calculator for a video game i dont wanna do the algebra from here 😭😭😭😭😭😭😭😭😭
thx anyway tho im prob gonna cook with this
oh nvm its kinda easy carry on
Unfortunately knowing algebra is a good basis from which we can learn many other things.
@@MichelvanBiezen i said i got it in fact imma start looking at algebra
You'll never regret that you took the time to learn algebra.
100th comment
We appreciate it! 🙂
What's the name of the last identity that he used, the one with the phase angle?
thanks sir for your this amazing guidance to Radha Singh from India
How about if both initial height and end height is not equal to zero ?
The problem would be worked exactly the same way. (The difference in height between the starting point and the end point is just a constant)