Question

A ball is dropped from a height of 60.0 m. A second ball is thrown from the height 0.850 s later. If both balls reach the ground at the same time, what was the initial velocity of the second ball?

Answer 1

find flight time of 1st ball ... 60.0 = 1/2 g T^2 ... T^2 = 120. / 9.8

find flight time of 2nd ball ... t = T - 0.850

0 = -1/2 g t^2 - v t + 60.0 ... solve for v (init velocity of 2nd ball)

Answer 2

First ball was "dropped" ,so height = -4.9t^2 + 60
time taken to hit the ground: 4.9t^2 = 60 ----> t = 3.49927...
The second ball took (3.49927 - .85) or 2.6493 seconds

If they reach the ground at the same time, the second ball must have had
some initial downwards velocity
height = -4.9t^2 + kt + 60 , (expect k to be negative)

height = 0 = -4.9(2.6493)^2 + k(2.6493) + 60
2.6493k = -25.6086..
k = -9.666 m/s <------- the initial velocity of the 2nd ball

better check my arithmetic

Answer 3

be aware of significant figures ... calculators are not necessarily impressive