A small object begins a free‑fall from a height of 25.0 m. After 1.30 s, a second small object is launched vertically upward from the ground with an initial velocity of 38.0 m/s. At what height ℎ abo

Question

A small object begins a free‑fall from a height of 25.0 m. After 1.30 s, a second small object is launched vertically upward from the ground with an initial velocity of 38.0 m/s.

At what height ℎ above the ground will the two objects first meet?

Answer 1

The height of the first object =25-.5gt². The height of the second object =38(t-1.3)-.5g(t-1.3)². Set these two expressions equal to each other and solve for t (3.28sec).

Answer 2

y₁ = 25 - gt²/2, y₂ = 38(t - 1.3) - g(t - 1.3)²/2;

y₁ = y₂; 25 - gt²/2 = 38t - 49.4 - gt²/2 + 1.3gt - 1.69g/2; 25 + 1.69·4.9 + 49.4 = (38 + 1.3·9.8)t; t = 1.6295 s

h = 25 - 9.8·1.6295²/2 = 12 m