Question

2. Derive an expression for the final velocity of an object dropped from some height (h)

above the ground, initially at rest, the instant before hits the ground. Your equation
should be in terms of (h) and (g) only​

Answer 1

v = √(2gh)

Step-by-step explanation:

Given:

Δy = h

v₀ = 0

a = g

Find: v

v² = v₀² + 2aΔy

v² = (0)² + 2gh

v = √(2gh)

You can also prove this with conservation of energy.

PE = KE

mgh = ½mv²

v = √(2gh)