How Is Average Force Calculated in Ball Bounce Dynamics?

[perfect_piccolo]

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?

 

[Doc Al]

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.

 

[perfect_piccolo]

ok thanks got it!

 

[perfect_piccolo]

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

$$\Sigma$$Fy2 = 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

 

[Doc Al]

It looks like you analyzed the vertical components, so now analyze the horizontal components.

 

[Saladsamurai]

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

$$\Sigma$$Fy2 = 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