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14-10-2020
Physics

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A ball is dropped from a 19m high cliff. The acceleration on the ball was 9.8m/s². What was the ball's final velocity before hitting the ground?

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14-10-2020

Answer:

19.3 m/s

Explanation:

Take down to be positive. Given:

Δy = 19 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) (19 m)

v = 19.3 m/s

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