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

Use the equations:
S=vi*t + (1/2)gt^2

For 1st ball:
20=0+(1/2)gt^2 => t=sqrt(40/g)

For 2nd ball:
20 = vi*(t-1) + (1/2)g(t-1)^2
vi = [20-(1/2)g(t-1)^2]/(t-1)
=(20-(9.81/2)*(t-1)^2)/(t-1)
=14.62 m/s downwards