posted by bethany
While standing on a bridge 15.0 m above the ground, you drop a stone from rest. When the stone has fallen 5.00 m, you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction.
First calculate the time to fall Y = 5.00 m and Y = 150 m for the first (dropped) stone. t = sqrt(2Y/g)
The times are 1.02 s and 5.53 s
Then calculate the speed V the second stone must be thrown to reach the ground at the same time.
150 = V*(t-1.02) + (g/2)(t-1.02)^2
Solve for V