Asked by Alex

A child riding a bicycle has a total mass of 40.0 kg. The child approaches the top of a frictionless hill that is 10.0 m high and 100.0 m long at 5.0 m/s. What is the child’s velocity at the bottom of the hill? b) (Disregard air resistance. g = 9.8 m/s2.)
Answered by oobleck
1/2 mv^2 = mgh
Answered by Dr Adi ([email protected])
According to the work energy theorem,
Initial (KE+PE) =Final (KE+PE)
0.5x40x5^2 + 40x9.8x10 = KE final + 0
500 + 3920 = 1/2 mv2^2
v2^2 = 2 x 4420 / 40
v2 = 14.87 m/s
Answered by adibhalekargmailcom
Frictionless hill is considered.
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