a stone is thrown straight upwards from the edge of a 24.0 m high cliff. It just misses the cliff on the way down, and it hits the ground below with a velocity of - 42.3 m/s. With what velocity was the stone thrown upward initially?

Question

Shayla Rich

Physics

25 October, 17:27

a stone is thrown straight upwards from the edge of a 24.0 m high cliff. It just misses the cliff on the way down, and it hits the ground below with a velocity of - 42.3 m/s. With what velocity was the stone thrown upward initially?

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Answers (1)

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Jovani Mcbride · Answer 1

47.5 m/s

Explanation:

Given:

Δy = - 24.0 m

v = - 42.3 m/s

a = - 9.8 m/s²

Find: v₀

v² = v₀² + 2aΔy

v² = (-42.3 m/s) ² + 2 (-9.8 m/s²) (-24.0 m)

v = 47.5 m/s