Physics
posted by Alex on .
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?

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*(t1) + (1/2)g(t1)^2
vi = [20(1/2)g(t1)^2]/(t1)
=(20(9.81/2)*(t1)^2)/(t1)
=14.62 m/s downwards