Question

A brick falls from rest from the roof of a tall building. Two seconds later a ball is thrown vertically downward with a velocity of 75m/s how far below the roof does the ball hit the brick?

Answer 1

Let us consider the downwards direction as positive.

Gravitational acceleration = g = 9.81 m/s²

Initial velocity of the brick = V₁ = 0 m/s

Initial velocity of the ball = V₂ = 75 m/s

Time interval between the brick and the ball = t = 2 sec

Total time period the brick travels before being hit by the ball = T₁

Total time period the ball travels before hitting the brick = T₂

T₁ = t + T₂

T₁ = 2 + T₂

Distance traveled by the brick before being hit by the ball = D₁

Distance traveled by the ball before hitting the brick = D₂

The distance traveled by the brick and the ball are the same therefore,

T₂ = 0.3543 sec

D₂ = 27.2 m

Distance below the roof the ball hits the brick = 27.2 m