How Is Average Force Calculated in Ball Bounce Dynamics?

AI Thread Summary
The discussion revolves around calculating the average force exerted on a ball during its bounce dynamics after falling from a height. The initial calculations provided a final velocity of 24.17 m/s before impact and a rebound velocity of 21.64 m/s, leading to confusion over the average acceleration and force. Participants emphasized the importance of correctly applying signs in the equations to determine the average acceleration and subsequently the force exerted by the ground. Additionally, a separate query regarding forces in a physics problem involving a leg and cast was introduced but suggested to be moved to a new thread for better engagement. The conversation highlights the complexities of force calculations in dynamics and the need for clarity in problem-solving approaches.
perfect_piccolo
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Homework Statement



After falling from rest at a height of 29.8 m, a 0.596 kg ball rebounds upward, reaching a height of 23.9 m. If the contact between ball and ground lasted 1.68 ms, what average force was exerted on the ball?


Homework Equations





The Attempt at a Solution



so I figured out that the final velocity before the ball first hits the ground is 24.17 m/s, and the velocity when the ball begins to bounce back up its inital velocity is 21.64 m/s. If the a= vf-vi / t, that gives me 21.64 - 24.17 which is -2.53 / 0.00168 which equals -1505.95, which I think is totally unreasonable...where am I going wrong?
 
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Careful with signs. Downward velocity should be negative. That will give the average acceleration; you still have to calculate the force of ground on ball.
 
ok thanks got it!
 
All right new question :D

The leg and cast in the figure below weigh 232 N (w1). (Assume w3 = 100 N and q = 39.2o). (See pic)


So I know that Ft3 = Fg3 = 100N

\SigmaFy2 = m2ay2
Ftyx + Fty3 - Fg1 = 0
Ft2 sin \alpha + Ft3sin\vartheta - Fg1 = 0
Ft2 sin \alpha +100sin39.2 - 232 N = 0
Ft2 sin \alpha = 168.797

But now I'm stuck...I would like to solve for Fty2, because I have found that Ftx2 is 77.49, but I'm not sure how to solve fot Fty2 with the information I have
 

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It looks like you analyzed the vertical components, so now analyze the horizontal components.
 
perfect_piccolo said:
All right new question :D

The leg and cast in the figure below weigh 232 N (w1). (Assume w3 = 100 N and q = 39.2o). (See pic)


So I know that Ft3 = Fg3 = 100N

\SigmaFy2 = m2ay2
Ftyx + Fty3 - Fg1 = 0
Ft2 sin \alpha + Ft3sin\vartheta - Fg1 = 0
Ft2 sin \alpha +100sin39.2 - 232 N = 0
Ft2 sin \alpha = 168.797

But now I'm stuck...I would like to solve for Fty2, because I have found that Ftx2 is 77.49, but I'm not sure how to solve fot Fty2 with the information I have



I would start a new thread for this, it will get much better traffic. Also, if you have an image host like Photobucket, I would upload though that. The PF image hosting has to be approved by a moderator first...which could take all day as they have a lot of them to deal with.

Casey

Edit: But it looks like Doc Al was just waiting to prove me wrong:rolleyes:
 
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