A stone falls freely from a height of 100 m. After 1 s, another stone is thrown vertically downward

A stone falls freely from a height of 100 m. After 1 s, another stone is thrown vertically downward from the same height. How fast is the second stone to be thrown so that both stones fall to the ground at the same time?

Given: H (initial height of falling stones) = 100 m; Δt (time difference) = 1 s; the fall of the first stone is free.

Reference values: g (acceleration due to gravity) ≈ 10 m / s2.

1) The duration of the fall of the first stone: H = g * t1 ^ 2/2; t1 = √ (2H / g) = √ (2 * 100/10) = 4.47 s.

2) The duration of the fall of the second stone: t2 = t1 – Δt = 4.47 – 1 = 3.47 s.

3) Initial speed of the second stone: H = V0 ^ 2 * t ^ 2 + g * t2 ^ 2/2; V02 = (H – g * t2 ^ 2/2) / t2 = (100 – 10 * 3.47 ^ 2/2) / 3.47 = 11.47 m / s.