Answer:
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
