Question

Jack drops a stone from rest off of the top of a bridge that is 24.4 m above the ground. After the stone falls 6.6 m, Jill throws a second stone straight down. Both rocks hit the water at the exact same time. What was the initial velocity of Jill's rock? Assume upward is the positive direction and downward is negative. (Indicate the direction with the sign of your answer.)

Answer 1

-17.555m/s

first I found the time it took for jacks stone to reach the bottom, using the formula vf = vi + at, vf and vi are final and initial velocities.

then i found the velocity at 6.6m using vf^2 = vi^2 + 2ad
and I found the time it took to get to 6.6m, so that I knew how long Jill waited to throw her stone, I used the formula d = t(vi+vf)/2, then i done total time - the time she waited, to get the time it took for there stones to hit the ground at the same time.

then to find the initial velocity of her throw I used the formula d = vit + (at^2)/2