If it wasn't for your videos, I would've never grasped the whole idea about projectile motion. You don't know how much help your videos have been to me! I am highly in debt and grateful! Thank You So much and Keep up the Enlightenment!
based on your all videos about projectile motion. I realised that you put either vertical/horizontal first on your working is because it depends on the time. If there is a time in the question, you would firstly use horizontal, vice versa.
@DeanNickChase If you look closely you can see in the diagram that it is a stick horse, ridden by a stick figure. Stick horses (and stick figures) are two-dimensional, so they are very thin and extremely lightweight. This gives the horse a tremendous power-to-weight ratio, allowing him to make the jump that a big fat three-dimensional horse could not make.
@derekowens as the previously mentioned teacher from TheJezza2, must say that I respect your work and clear explanations. I have commonly referenced your videos in my physics class for at home help. Keep doing the good "f x d x Cos(theta)"!
@getrichquickideas Thinking from the start to the peak, which is only half of the flight, gives you a result of 2.5 seconds. This is half the total time, so the total time is five seconds.
@H2daH acceleration is zero horizontally. This is always the case for projectiles because gravity always pulls down, and never exerts any force horizontally.
hey there, I really appreciate your video. but at 3:25 I did not really understand much. Why would you make the time 2.5s? I know that at 2.5s that's when farmer bob and his horse reached maximum height, but why would you do that? Is there another way to get the initial vertical velocity here? thanks
@jkoscak I prefer m/s, although m s^-1 is mathematically equivalent, and can fit neatly on one line when typing. I think the algebra is more clear, though, with it written as a fraction.
what is the reason we should assume that there is no horizontal acceleration? WHat if you jump from a stop? there would be both horizontal and veritcal acceleration
+xPinoyTribal Right, I see what you are saying. While you are jumping, pushing with your legs, there is definitely horizontal as well as vertical acceleration. Once you are airborne, however, coasting and out of contact with the ground, then at that point you are properly considered a projectile, and the only acceleration is the downward acceleration of gravity (assuming that we ignore air resistance).
+xPinoyTribal pre hehe, kaya walang horinzontal ang acceleration, kasi unbalance ang force sa x-direction(horizontal) dahil narin sa air resistance, may ang acceleration sa Y vertical acceleration dahil constant at uniform ang force, at ang tawag don gravitational force 9.81m/s so dahil nag rereact lang ang gravity sa vertical , ang Y lang may acceleration kasi nag babago ang acceleration dpende sa time at distance, habang constant palagi ang horizontal, gets mo po ?> hehehe
9.8 m/s^2 is the acceleration due to gravity. At least that's the value near earth's surface. And since a lot of physics problems take place near earth's surface, that value shows up fairly often, so it's a good number to know for basic physics work.
i didn’t consider the “start from peak” part for the vertical component solving, but i still got 24.5 seconds as the answer,,, is that alright?? THANK YOUU
Yes, there are other ways to solve it. You can consider the motion from start to finish, and as long you use the correct initial and final velocity you should get the same answer for the time.
@@derekowens since the flight time is 5 sec which means at t equal 5 they reached the ground doesn't that mean that at t equal 5 the vertical velocity equal 0?
@@adnankamal3697 t=5s is the end of the flight, so the velocity at t=5s is the impact velocity. Think of the final velocity as being the speed with which it hits.
this is probably too late but here lol when a object travels in a parabolic path, there comes a time ,right before when the object starts going down, the object only has horizontal force acting on it as the vertical force in that very instance is zero(the object goes up and when its done going up, in that very moment the final velocity is 0), and since its a parabola that means it took 2.5 seconds to go up and 2.5 to go down
Tangent theta is the ratio of opposite over adjacent but we don't want tan theta we want theta so when we take tan to the other side of the equation we get tan inverse theta (oppo/adj). I hope that was helpful :)
Thats half of the time so the velocity in the y direction is changing from up to down which makes the velocity in the y-direction 0, which enables you to plug the numbers in to solve for the inital velocity of y.
well here's another approach: let T = total time (start to peak, peak to...) T = 5s. R = 100m. ¢ = angle since T = 2Voy/g T = 2Vosin(¢)/g Vo = gT/2sin(¢) --------- eq1. Vo = (9.81)(5)/2sin(¢) x = Voxt ---------------------- eq2 100 = Vocos(¢)(5) Vo = 100/5cos(¢) equate e1 & eq 2 9.81(5)/2sin(¢) = 100/(5cos(¢)) sin(¢)/cos(¢) = 5(5)(9.81)/2(100) tan(¢) = {(5)(5)(9.81)/2(100)} ¢ = 50.81° substituting ¢ to either eq1 or eq.2 Vo = 31.6 m/s
![[UNCUT] ลูกหมี-ลิลลี่ เหงียน เปิดใจฟาด ปู มัณฑนา I คนดังนั่งเคลียร์ I 15 ก.ค.67](http://i.ytimg.com/vi/FKQ60Y7bJ6s/mqdefault.jpg)
If it wasn't for your videos, I would've never grasped the whole idea about projectile motion. You don't know how much help your videos have been to me! I am highly in debt and grateful! Thank You So much and Keep up the Enlightenment!
based on your all videos about projectile motion. I realised that you put either vertical/horizontal first on your working is because it depends on the time. If there is a time in the question, you would firstly use horizontal, vice versa.
thank you so much for this! it's so confusing when my teacher does it but you explain it so well!
Your videos are incredible. You sir are a miracle worker. thank you very much
@DeanNickChase If you look closely you can see in the diagram that it is a stick horse, ridden by a stick figure. Stick horses (and stick figures) are two-dimensional, so they are very thin and extremely lightweight. This gives the horse a tremendous power-to-weight ratio, allowing him to make the jump that a big fat three-dimensional horse could not make.
you're doing my professor's job for him. thank you
@derekowens as the previously mentioned teacher from TheJezza2, must say that I respect your work and clear explanations. I have commonly referenced your videos in my physics class for at home help. Keep doing the good "f x d x Cos(theta)"!
i love that drawing skills :D
Thanks a bunch its been long years since I was in physics
Best video out there!
ur drawings make physics much more enjoyable :)
Your videos are incredibly helpful! Thanks. :-)
100m??!!!! I need to get me that horsey!!!
Great video, Derek. Thank you for breaking them down in steps.
Ahhhh yeeeah! Bring it on exam 1. Thanks so much mista! :)
Thanks for the vids! They really helped me alot. :D
YOUR GREAT , IF U ARE MY PHYSICS TEACHER SWEAR TO GOD I PASSED WITH HONOR
Its really saddening to see the view count of such a magnificent video...
Thanks for these amazing videos, now when i solve a problem i can actually say that i understand whats going on haha
@getrichquickideas Thinking from the start to the peak, which is only half of the flight, gives you a result of 2.5 seconds. This is half the total time, so the total time is five seconds.
@H2daH acceleration is zero horizontally. This is always the case for projectiles because gravity always pulls down, and never exerts any force horizontally.
wow, great job explaining! very helpful
your such a good teacher derek :D
ok thanks derek for this video.
thank you very much sir!.keep it up :D you are a much better than my prof.haha
It works. Thank you
thank you very much.
This is intresting. VERY.
Dude, you are the best
u got me!!!! good one thx
to find the initial velocity..can i use the parrelogram law/triangle law insead of pythogorus theorem
hey there, I really appreciate your video. but at 3:25 I did not really understand much. Why would you make the time 2.5s? I know that at 2.5s that's when farmer bob and his horse reached maximum height, but why would you do that? Is there another way to get the initial vertical velocity here? thanks
@jkoscak I prefer m/s, although m s^-1 is mathematically equivalent, and can fit neatly on one line when typing. I think the algebra is more clear, though, with it written as a fraction.
these videos are fucking awesome!! i'm gonna go ham on my test on monday!!
Thank you so much.
This guy's drawing is awesome
This is awesome
what if it asks for the landing velocity? would that just be the horizontal velocity, which is 20m/s?
please answer my question, thank you!
@emiri, thank you.
Isnt it easier to use U for initial velocity instead of V zero?
Thank you 🙏
thank you so much
Why is the time for horizontal (5sec) different from you time for the verticle (2.5 sec) ?
i'm sorry. but, shouldn't be the initial velocity for x Vocos(theta)? thus, Vo=100m/(5s cos(theta))?
thank you!!!!!
im here 14 years later
And I'm still here, too!
Well done
what is the reason we should assume that there is no horizontal acceleration? WHat if you jump from a stop? there would be both horizontal and veritcal acceleration
+xPinoyTribal Right, I see what you are saying. While you are jumping, pushing with your legs, there is definitely horizontal as well as vertical acceleration. Once you are airborne, however, coasting and out of contact with the ground, then at that point you are properly considered a projectile, and the only acceleration is the downward acceleration of gravity (assuming that we ignore air resistance).
+xPinoyTribal pre hehe, kaya walang horinzontal ang acceleration, kasi unbalance ang force sa x-direction(horizontal) dahil narin sa air resistance,
may ang acceleration sa Y vertical acceleration dahil constant at uniform ang force, at ang tawag don gravitational force 9.81m/s so dahil nag rereact lang ang gravity sa vertical , ang Y lang may acceleration kasi nag babago ang acceleration dpende sa time at distance, habang constant palagi ang horizontal, gets mo po ?> hehehe
Thank you so much
the lesson here is clear. on my channel is a projectile motion demo that you can watch. it's actually a program that you can download from softpedia
Whats up Mr.Butler's class!!!
what software do you use?
why dont you use 1/2 as often found in the formula 1/2at squared. When you calculate V zero
why is (a)=0 ?
in the first horizontal eqaution
The only reason I'm watching this is because it's a physics homework assignment.
Doesn't tan^-1(24.5/20)= 56.4 degrees?
m/s or ms-1 ? what do you like better? :p
great
Can you lmk where the 9.8m2 come from
9.8 m/s^2 is the acceleration due to gravity. At least that's the value near earth's surface. And since a lot of physics problems take place near earth's surface, that value shows up fairly often, so it's a good number to know for basic physics work.
i didn’t consider the “start from peak” part for the vertical component solving, but i still got 24.5 seconds as the answer,,, is that alright?? THANK YOUU
oh wow didn’t notice this video was posted 11 years ago already oh wow
Yes, there are other ways to solve it. You can consider the motion from start to finish, and as long you use the correct initial and final velocity you should get the same answer for the time.
@@boss2dawerld750 11 years, and I'm still here! Lots to be thankful for.
@@derekowens since the flight time is 5 sec which means at t equal 5 they reached the ground doesn't that mean that at t equal 5 the vertical velocity equal 0?
@@adnankamal3697 t=5s is the end of the flight, so the velocity at t=5s is the impact velocity. Think of the final velocity as being the speed with which it hits.
Bravo this video is very good keep up the good work.
Would you explain why in vertical initial velocity, v=0? Great video. Thanks
this is probably too late but here lol when a object travels in a parabolic path, there comes a time ,right before when the object starts going down, the object only has horizontal force acting on it as the vertical force in that very instance is zero(the object goes up and when its done going up, in that very moment the final velocity is 0), and since its a parabola that means it took 2.5 seconds to go up and 2.5 to go down
where did 9.8 m/s come from??
vertical acceleration from gravity
How did you get the number 9.8 m/^2?
because of gravity
yo, i can see u r a good drawer haha
@smileinc oh nevermind, my calculator needed reseting. :(
is the peak half of the total time?
+irwin israel Tomas Yes, but only if it starts and ends at the same height.
The problem is quite old.
So the video is.
Can someone explain to me why he used inverse tangent?
Tangent theta is the ratio of opposite over adjacent but we don't want tan theta we want theta so when we take tan to the other side of the equation we get tan inverse theta (oppo/adj). I hope that was helpful :)
Use the inverse tangent to determine angle in degrees and seconds
Oh, I took this class in the summer, but thanks anyways guys.
thnks
I'm watching this video in 2023
👍
When your doing the vertical motion, why is time 2.5 seconds? Please answer...
Thats half of the time so the velocity in the y direction is changing from up to down which makes the velocity in the y-direction 0, which enables you to plug the numbers in to solve for the inital velocity of y.
well here's another approach:
let T = total time (start to peak, peak to...)
T = 5s.
R = 100m.
¢ = angle
since T = 2Voy/g
T = 2Vosin(¢)/g
Vo = gT/2sin(¢) --------- eq1.
Vo = (9.81)(5)/2sin(¢)
x = Voxt ---------------------- eq2
100 = Vocos(¢)(5)
Vo = 100/5cos(¢)
equate e1 & eq 2
9.81(5)/2sin(¢) = 100/(5cos(¢))
sin(¢)/cos(¢) = 5(5)(9.81)/2(100)
tan(¢) = {(5)(5)(9.81)/2(100)}
¢ = 50.81°
substituting ¢ to either eq1 or eq.2
Vo = 31.6 m/s
U did all that for no reply
well, u replied
thank you
wow! I am very impressed by your work. well done
Laughing at the reply to the horse comment.
Easy