- 5

- 0

**1. Homework Statement**

a ball is thrown vertically upward with an initial speed of 19 m/s. Then, .97 s later, a stone is thrown straight up (from the same inital height as the ball) with an inital speed of 31 m/s. How far above the release point will the ball and stone pass each other. Answer is units of m. Gravity is -9.8 m/s^2.

**2. Homework Equations**

I tried using v^2=v0^2*2ax and x=x0+v0t+0.5at^2

**3. The Attempt at a Solution**

I solved 19(.97)+0.5(-9.8)(.97)^2 and got 13.82 m for the ball after .97 s.

After this I got lost and didn't know what to do. I know this means that when the ball is at 13.82 m/s that the stone is just being thrown. Please help!!