A stone of mass \'m\' is dropped from rest at a height of 1.58 m. From what heig

Question

A stone of mass 'm' is dropped from rest at a height of 1.58 m. From what height would a stone of mass 'm/2' have to be dropped to have the same momentum upon striking the ground?

Explanation / Answer

We know that when stone falling from rest we have            h= 1/2gt2 ==> 1.58 = 4.9t2 ==> t = 0.5678s So that v = at = 9.8 * 0.5678 = 5.564 m/s momentum = mv = (m)* 5.564 = 5.564m Now with a mass of m/2 having a momentum of 5.564m we have todouble the velocity. 5.562m = 5.564(m/2)*2 = 11.129m To get a velocity of 11.129m/s 11.129 = gt so t= 1.135 s Therefore the distance is h' =1/2gt2                                          = 4.9 * 1.135 * 1.135                                          = 6.32 m

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