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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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  1. 25 October, 17:42
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    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
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