Question

An object is dropped at a height of 6 m from the ground. How fast is it moving just before it hits the ground?

Answer 1

Step-by-step explanation:

Given:

Δy = 6 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (6 m)

v ≈ 10.8 m/s

Round as needed.