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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)