A person jumps from the roof of a house 5.5 m high. When he strikes the ground b

Question

A person jumps from the roof of a house 5.5 m high. When he strikes the ground below, he bendshis knees so that his torso decelerates over an approximatedistance of 0.60 m. The mass of his torso(excluding legs) is 41 kg. (a) What is his velocity just before his feetstrike the ground?
1 m/s (downward)

(b) What is the average force exerted on his torso by his legsduring deceleration?
2 N (upward) (a) What is his velocity just before his feetstrike the ground?
1 m/s (downward)

(b) What is the average force exerted on his torso by his legsduring deceleration?
2 N (upward)

Explanation / Answer

a ) Height (h)= 5.5 m     acceleration(a) = 9.8 m/s 2     initial velocity(u) = 0     using the equation     v 2- u2 = 2aS     Final velocity (v)=2aS=(2*9.8*5.5) = 10.38 m/s b ) At the time of stopping ,    total change in momentum = mass* (Final - initialvelocity ) = 41 * (10.38 - 0 ) = 425.7 kgm/s This should be equal to Force* time    (Force = change in momentum / time ) So F * 0.60 = 425 . 7      F = 709.48 N

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