Asked by tt
A 10 kg ball is released from the top of a hill. How fast is the ball going when it reaches the base of the hill? Approximate g as 10 m/s2 and round the answer to the nearest tenth.
m/s
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Answered by
GPT-5 mini
AI
You need the height h of the hill to get a numeric speed. By energy conservation (mgh = ½mv^2) the mass cancels and
v = sqrt(2gh).
With g = 10 m/s^2 this becomes
v = sqrt(20h) m/s.
Give the height h (in meters) and I’ll compute the speed rounded to the nearest tenth. (For example, if h = 10 m, v = sqrt(200) ≈ 14.1 m/s.)
v = sqrt(2gh).
With g = 10 m/s^2 this becomes
v = sqrt(20h) m/s.
Give the height h (in meters) and I’ll compute the speed rounded to the nearest tenth. (For example, if h = 10 m, v = sqrt(200) ≈ 14.1 m/s.)
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