# Kinematics free fall problems part 3 help

1. May 10, 2010

### P944

1. The problem statement, all variables and given/known data
To find the height of an overhead power line, you throw a ball straight upward. The ball passes the line on the way up after 0.85 s, and passes it again on the way down 1.2 s after it was tossed.

(a) What is the height of the power line?

(b) What is the initial speed of the ball?

2. Relevant equations
v = v0 + at
x=x0+v0*t + 1/2*a*t^2
x = x0 +v*t - 1/2*a*t^2
v^2 = v0^2 +2*a*(x-x0)

3. The attempt at a solution

I drew a picture and decided to solve for intial speed first. I used the equation v = v0 + a* t . V=0, a = -9.8m/s^2 and i took time at v0 to be 1.25 + 0.85/2 = 1.05 s
V0 = 10.29m/s

I then solved part a by using V^2 = V0^2 + 2 *a*(x-x0)
0=(10.29)^2 +2(9.8)(x-0)
i solved this and ended up with 5.40m

I got both of these wrong on my practice test. I would greatly appreciate if someone could explain to me what i did wrong. Thanks so much!

2. May 10, 2010

### Char. Limit

3. May 10, 2010

### P944

oh you are correct! i'm dyslexic! :( I resolved the problem and got 10.04m/s for my initial speed which is correct. However, my distance of 5.148m is still wrong. Any ideas?

4. May 10, 2010

### Cyosis

In your equation you set v equal to 0. This only happens at one point which is when the ball reaches its maximum height. Nowhere is it stated that the power line is at the top of the ball's trajectory. In fact it is clearly stated that the ball passes the power line twice, which can only happen if the power line is below the ball's maximum height.

5. May 10, 2010

### P944

you are correct, now that i got my initial velocity of 10.04m/s, g = -9.80m/s^2, x0= 0 and time crossing power line =0.85 i used x = x0 + V0 * t + 1/2*g*t^2
x = 0 + (10.04m/s)(0.85s ) + 1/2 (-9.8m/s^2)(0.85)^2 = 4.994 m How does that look?