While standing on a bridge 15 m above the ground, you drop a stone from rest. When the stone has fallen 3.2 m, you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant?

11.3 m/s

Explanation:

First, find the time it takes for the first stone to fall 3.2 m.

Given:

Δy = 3.2 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(3.2 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.81 s

Next, find the time for the first stone to land.

Given:

Δy = 15 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(15 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 1.75 s

The difference in time is 1.75 s - 0.81 s = 0.94 s. Find the initial velocity needed for the second stone to land after that amount of time.

Given:

Δy = 15 m

a = 9.8 m/s²

t = 0.94 s

Find: v₀

Δy = v₀ t + ½ at²

(15 m) = v₀ (0.94 s) + ½ (9.8 m/s²) (0.94 s) ²

v₀ = 11.3 m/s