A ball is dropped from rest from a height of 20.0m. One second later a second ball is thrown vertically downwards. If the two balls arrive on the ground at the same time, what must have been the initial velocity of the second ball?

Answer 1

First Ball.
d = Vo*t + 0.5g*t^2 = 20 m.
0 + 4.9t^2 = 20
t^2 = 4.08
Tf = 2.02 s. = Fall time

2nd Ball.
d = Vo*t + 0.5g*T^2 = 20.
Vo*(2.02-1) * 4.9*(2.02-1)^2 = 20
1.02Vo + 4.9*(1.02)^2 = 20
1.02Vo + 5.1 = 20
1.02Vo = 20 - 5.1 = 14.9
Vo = 14.6 m/s.