Jumping to the ground. A 75.00-kg man steps off a platform 3.10 m above the grou
ID: 1763349 • Letter: J
Question
Jumping to the ground. A 75.00-kg man steps off a platform 3.10 m 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 .60 m before coming to rest. a) what is his speed at the instant his feet touch the ground? b) Treating him as a particle, what is his acceleration (magnitude and direction) as he slow down, if the acceleration is assumed to be constant? c) draw his free-body diagram, in terms of forces on the diagram, what is the net force on him? Use Newton's laws and the results of part b) to calculate the average force his feet exert on the ground while he slows down. express this force in newtons and also as a multiple of his weight. Please walk me step!Explanation / Answer
(a) the speed at the time his feet touch the ground v = (2gh) = (2 * 9.8 * 3.10) = 7.8 m/s (b) As he slows down, lets assume his deceleration is a v2 - u2 = 2as a = - u2 / 2s = - 7.8 *7.8 / 2*0.60 = - 50.7m/s2 It sign is negative, because it is opposing themotion. (c) The time taken to come to rest v = u - at 0 = 7.8 - 50.7 *t => t = 0.15 s According to Newton's law we have Impulse = change in momentum F*t = m * u F * 0.15 = 75 * 7.8 F = 3900 N
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