Question

If a bowling ball ball was released from the 6th floor of a hotel room 22 meters from the ground, at what velocity would the ball be moving just before it hits the head of the man who is 1.5 meters tall? Solve using kinematic equations, unknown values, and SOHCAHTOA.

Answer 1

Explanation:

Given:

y₀ = 22 m

y = 1.5 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2 (-9.8 m/s²) (1.5 m − 22 m)

v = -20.0 m/s