I would explain that the equation for the height is a parabola. So there are two solutions: one for going up and going down for the 10 ft mark. However, the velocity has a negative sign once starts going down. Once you calculate the time for 15 ft, you define the location. I guess, at least for me if you see the description of the trajectory as a parabola, it is easier to understand.
Velocity is not a force. This analysis starts from the instant that the ball leaves the hand. Once the ball doesn’t touch the hand the only force acting on the ball is gravity (ignoring air resistance). It will be the hand that produces the initial force on the ball to cause the movement.
hi sir can we cal calculate initial velocity v by using formula R = v2 sin(2theta)/ g R = 15 feet = 4.57 m theta = 45 degree if yes then im getting v as 6.69 m/s pls clearify
Oh wait I have another question sir. In 12:59 you showed how to solve the time it takes for the ball to reach the rim, and actually we are doing our research now and we know the t or the time it takes for the ball to pass through the rim. So, can we simply calculate the initial velocity from there like: Vo = L/((cos0)t) ?
Hi Physics Ninja! It was a very informative video! However, I have a question. Is it possible to calculate the force exerted by the player (in your example) without the acceleration and final velocity of the ball given?
You lost me at 15:40 I don't understand how all the v sin thetas (I count four of them), only turn into v^2sin theta^2 -- it looks like either half of the equation would be that, but one just disappears. I can't base my studying on terms seemingly disappearing, so I can't use this to study, even if the answers are right. Thanks for trying.
Last velocity value -3.18 is actually your value in ft/sec not m/s. Your value in m/s should be −0.99584 m/s. If you us this the angle should be actually -10.5 degree thus 10.5 down from the horizontal. Cheers and thanks you have fun content...
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I would explain that the equation for the height is a parabola. So there are two solutions: one for going up and going down for the 10 ft mark. However, the velocity has a negative sign once starts going down. Once you calculate the time for 15 ft, you define the location. I guess, at least for me if you see the description of the trajectory as a parabola, it is easier to understand.
Glad you are back. Haven't seen you in 4 months xD
I took the summer off but am back now. Looking to make 1-2 videos/week. Thanks for following Physics Ninja.
is the velocity you found for the final question the same as the final velocity?
you are doing great
Yayyy tysm!! I really needed this!
How could the height of 3 meters and the acceleration due to gravity both be positive
May I ask why the Vy= Voy-gt is not Vy=Voy-gt because the Vo of the Vy would be zero cuz it's from the max height?
Thank you!
sir why did you used 9.81 m/s than 32.2 ft/s even though its feet?
Hello excuse me, is the velocity considere as a force? Otherwise I want to know how the movement on the horizontal direction is made.
Velocity is not a force. This analysis starts from the instant that the ball leaves the hand. Once the ball doesn’t touch the hand the only force acting on the ball is gravity (ignoring air resistance). It will be the hand that produces the initial force on the ball to cause the movement.
What Math is used in question 2?
good work
how would i solve this problem if my problem did not give me and initial height of ball but i have the initial velocity?
Thank you
hi sir
can we cal calculate initial velocity v by using formula R = v2 sin(2theta)/ g
R = 15 feet = 4.57 m
theta = 45 degree
if yes then im getting v as 6.69 m/s
pls clearify
Oh wait I have another question sir. In 12:59 you showed how to solve the time it takes for the ball to reach the rim, and actually we are doing our research now and we know the t or the time it takes for the ball to pass through the rim. So, can we simply calculate the initial velocity from there like:
Vo = L/((cos0)t) ?
Yes
Hi, I worked out 3.56 m for the maximum height.
Hi Physics Ninja! It was a very informative video! However, I have a question. Is it possible to calculate the force exerted by the player (in your example) without the acceleration and final velocity of the ball given?
Not possible
Ok thank you!
Btw how do you calculate the release force when shooting the basketball? Like are there any formulas for it?
@@ribandrey6430 I think for a basketball player, they could measure their max height, where Velocity Y =0 of their throw.
ITS NOT STRAIT forward to the point but its fine i learn something from it
What if you don’t know the initial angle and what to find it
th-cam.com/video/4jNE3eTVEgo/w-d-xo.html
You lost me at 15:40
I don't understand how all the v sin thetas (I count four of them), only turn into v^2sin theta^2 -- it looks like either half of the equation would be that, but one just disappears. I can't base my studying on terms seemingly disappearing, so I can't use this to study, even if the answers are right. Thanks for trying.
in 10:56 you said the height was 3 ft... isnt it 7ft which is 2.13 m ?!
Why did you take long gap bro...
Last velocity value -3.18 is actually your value in ft/sec not m/s. Your value in m/s should be −0.99584 m/s. If you us this the angle should be actually -10.5 degree thus 10.5 down from the horizontal. Cheers and thanks you have fun content...
God sent
I don't think the maximum height is correct. how would it be less than the height of the basketball rim? I got 3.56 m for ymax
10ft rim is 3.05m… all good
wowowwww.but how can i get acc ?
❤
😎
You remind me of Ben Affleck lol
what da fuck
you are wrong. sin 45 = cos45 that ridiculus.
It is correct