A 75.0 kg stunt man jumps from a balcony and falls 23.5 m before colliding with

Question

A 75.0 kg stunt man jumps from a balcony and falls 23.5 m before colliding with a pile of mattresses. If the mattresses are compressed 1.10 m before he is brought to rest, what is the average force exerted by the mattresses on the stuntman?

Explanation / Answer

average force (to stop motion) = change in momentum/time Fav (motion) = m[v - u] / delta t = - mu /delta t (final v=0) Fav (motion) = - mu /(displacement/average speed) Fav (motion) = - mu * (average speed)/d Fav (motion) = - mu * (u+0)/2/d Fav (motion) = - 0.5 mu^2 /d = decrease in KE/ d ------------------------------------ at the free surface of mattress>>> loss in PE (man) = gain in KE 0.5 m[u^2 - o] = mg*28 0.5 mu^2=75*9.81*28 now the mattress comes into picture, and this KE is made zero with d=1.2 m ------------------------- Fav (motion) = 75*9.81*28 /1.2 [upwards] ========================= this is the force mattress has to apply for countering motion downwards, but to support this man, its weight also need to supported (mg), which acts downwards net force (mattress, up) = [75*9.81*28 /1.2] + mg net force (mattress, up) = [75*9.81*28 /1.2] + [mg*d/d] net force (mattress, up) = [75*9.81*28 + mgd] /1.2 F =75*9.81[28 + 1.2] /1.2 F = 21483.9 / 1.2 = 17903. 25 N ========================= if you take g = 9.8 then F = 17885 N

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