Question

A large rock is released from rest from the top of a tall building. The average speed of the rock during the first second of the fall is 5 m/s. What is the average speed of the rock during the next second? (In this question we use the approximate value of 10 m/s² for the gravitational acceleration.)

Answer 1

Let us assume the downward direction as positive.

Gravitational acceleration = g = 10 m/s²

Initial velocity of the rock = V₀ = 0 m/s (Released from rest)

Velocity of the rock at the end of the first second = V₁

Time period = T₁ = 1 sec

V₁ = V₀ + gT₁

V₁ = 0 + (10)(1)

V₁ = 10 m/s

Distance traveled by the rock in the 2^nd second = D₂

Time interval = t = 2 - 1 = 1 sec

D₂ = V₁t + gt²/2

D₂ = (10)(1) + (10)(1)²/2

D₂ = 15 m

Average speed of the rock during the 2^nd second = V_2avg

V_2avg = 15 m/s

Average speed of the rock during the 2^nd second = 15 m/s