A boy of mass 65.6 kg is rescued from a hotel fire by leaping into a firefighter
ID: 1972231 • Letter: A
Question
A boy of mass 65.6 kg is rescued from a hotel fire by leaping into a firefighters' net. The window from which he leapt was 7.3 m above the net. The firefighters lower their arms as he lands in the net so that he is brought to a complete stop in a time of 0.40 s. Ignore air resistance.(a) What is his change in momentum during this time interval?
? kg·m/s upward or downward?
(b) What is the impulse on the net due to the boy during the interval? [Hint: Do not ignore gravity.]
? N · s upward or downward?
(c) What is the average force on the net due to the boy during the interval?
? N upward or downward?
Explanation / Answer
SOLUTION: (a) Mass of the boy, m = 65.6 kg window from which he leapt was above the net, h=7.3 m He is brought to a complete stop in a time of,t= 0.40 s Assume, initial speed of the mass, vo= 0 m/s Let, speed of the mass after falling a distance h is, v By using the kinematic equations , v2 - v02 = 2ah v2 - 02 = 2ah velocity of the mass, v = 2gh = [(2)(9.8 m/s2)(7.3 m)] = 11.96 m/s Hence, change in momentum, P = m(v - v0) = m (v - 0) = mv = (65.6 kg)(11.96 m/s) = 784.576 kg.m/s (down ward) _________________________________________________________ _________________________________________________________ _________________________________________________________ _________________________________________________________ (b) Therefore , impulse J = change in momentum = 784.576 kg.m/s (down ward)_________________________________________________________ _________________________________________________________ (c) Aerage force on the net is F = P/t = 784.576 kg.m/s / (0.4 s) = 1961.44 N (up ward)
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