Question

A stone is thrown vertically upwards from the ground with a velocity 15 m/s. At the same instant a ball is dropped from a point directly above the stone from a height of 30 m. At what height from the ground will the stone and the ball meet and after how much time? (Use g = 10 m/s² for ease of calculation).

Answer 1

Let the stone and the ball meet after time t₀. From second equation of motion, the distances travelled by the stone and the ball in that time is given as,

S_stone = 15 t₀ – \(\frac{1}{2}\) gt₀²

S_ball = \(\frac{1}{2}\) gt₀²

When they meet. S_stone + S_ball = 30

∴ 15t₀ – \(\frac{1}{2}\) gt₀² + \(\frac{1}{2}\) gt₀² = 30

t₀ = \(\frac{30}{15}\) = 2 s

∴ S_stone = 15 (2) – \(\frac{1}{2}\) (10) (2)² = 30 – 20 = 10 m

The stone and the ball meet at a height of 10 m after time 2s.