A stone is dropped from a 75-m - high building. When this stone has dropped 15 m, a second stone is thrown downward with an initial velocity such that the two stones hit the ground at the same time. What was the initial velocity of the second stone?

The initial velocity of the second stone is 34.3 m/s

Explanation:

The distance height covered by the second stone will be

75m - 15m = 60 m

Since the two stones hit the ground at the same time.

Using third equation of motion, which state that;

V^2 = U^2 - 2gH

Where

U = Initial velocity

V = final velocity

H = 60 m

g = 9.81 m/s^2

As the stone reaching the ground, that is, coming to rest, V = 0. Therefore,

Substitute all the parameters into the equation

0 = U^2 - 2 * 9.81 * 60

U^2 = 1177.2

U = 34.3 m/s

The initial velocity of the second stone is 34.3 m/s