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A 10 kg ball is released from the top of a hill measuring 2 m high. How fast is the ball going when it reaches the base of the hill at .5 m approximate G as 10 m/s squared and round the answer to the nearest 10th.

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MathPhys

Answer:

5.5 m/s

Explanation:

Initial potential energy = final potential energy + kinetic energy

PEâ‚€ = PE + KE

mgh₀ = mgh + ½ mv²

gh₀ = gh + ½ v²

v² = 2g (h₀ − h)

Plug in values:

v² = 2 (10 m/s²) (2 m − 0.5 m)

v² = 30 m²/s²

v ≈ 5.5 m/s

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