# Newton's Second Law Equations

1. Oct 12, 2007

### perfect_piccolo

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. 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?

2. Oct 12, 2007

### Staff: Mentor

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.

3. Oct 12, 2007

### perfect_piccolo

ok thanks got it!

4. Oct 12, 2007

### 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

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5. Oct 12, 2007

### Staff: Mentor

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

6. Oct 12, 2007