Question

A stone is dropped into a river from a bridge 47.8 m above the water. Another stone is thrown vertically down 1.60 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?

Answer 1

Let us assume the downwards direction as positive.

Gravitational acceleration = g = 9.81 m/s²

Height of the bridge = H = 47.8 m

Initial velocity of the first stone = V₁ = 0 m/s

Time taken by the first stone to hit the water = T₁

H = V₁T₁ + gT₁²/2

47.8 = (0)T₁ + (9.81)T₁²/2

T₁ = 3.122 sec

Initial velocity of the second stone = V₂

Time taken by the second stone to hit the water = T₂

The second stone is thrown 1.6 sec after the first stone and both the stones hit the water at the same time.

T₂ = T₁ - 1.6

T₂ = 3.122 - 1.6

T₂ = 1.522 sec

H = V₂T₂ + gT₂²/2

47.8 = V₂(1.522) + (9.81)(1.522)²/2

V₂ = 23.94 m/s

Initial speed of the second stone = 23.94 m/s