Question

A dynamite blast at a quarry launches a rock straight upward, and 1.8 s later it is rising at a rate of 12 m/s. Assuming air resistance has no effect on the rock, calculate its speed (a) at launch and (b) 5.4 s after launch.

Answer 1

Let us consider upwards as positive and downwards as negative.

Gravitational acceleration = g = -9.81 m/s²

Velocity of the rock at launch = V₀

Velocity of the rock 1.8 sec after launch = V₁ = 12 m/s

T₁ = 1.8 sec

V₁ = V₀ + gT₁

12 = V₀ + (-9.81)(1.8)

V₀ = 29.658 m/s

Velocity of the rock 5.4 sec after launch = V₂

V₂ = V₀ + gT₂

V₂ = 29.658 + (-9.81)(5.4)

V₂ = -23.316 m/s

The negative sign indicates the rock is moving downwards.

We are asked for the speed (magnitude) of the rock therefore it will be positive.

a) Speed of the rock at launch = 29.658 m/s

b) Speed of the rock 5.4 sec after launch = 23.316 m/s