A 72.4 man steps off a platform 3.02 above the ground. he keeps his legs straigh

Question

A 72.4 man steps off a platform 3.02 above the ground. he keeps his legs straight as he falls, but at the moment his feet touch the ground his knees begin to bend, and, treated as a particle, he moves an additional 0.620 before coming to rest

A.What is his speed at the instant his feet touch the ground? v=7.67m/s


B.Treating him as a particle, what is the magnitude of his acceleration as he slows down if the acceleration is constant? a=47.7m/s^2

C. Use Newton's laws and the results of part B to calculate the force the ground exerts on him while he is slowing down. Express this force in newtons. F=4170N

D.Use Newton's laws and the results of part to calculate the force the ground exerts on him while he is slowing down. Express this force as a multiple of the man's weight. F=5.87w

E.What is the magnitude of the reaction force to the force you found in part C

Explanation / Answer

(a) v^2 = u^2 +2gh here u = 0 Hence the speed at the time his feet touch the ground v = sqrt (2gh) = sqrt (2 * 9.8 * 3.10) = 7.8 m/s (b) As he slows down, lets assume his deceleration is a here u = 7.8 m/s a = ? v = 0 as he will halt and s = 0.60 m v^2 = u^2 - 2 a s => 0 = 7.8^2 - 2 a * 0.60 => a = 50.7 m/ss it sign will be negative as it is opposing the motion. (c) Newton's law say that Impulse = change in momentum ------ eqn 1 First find out time to come to rest => v = u - at => 0 = 7.8 - 50.7 * t => t = 0.15 s so according to eqn 1 Impulse = F * t = m * u F * 0.15 = 75 * 7.8 => average force F = 3900 N or = 3900/10 kgF I hope it helps.

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