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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?

Respuesta :

MathPhys

Answer:

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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