A stone is dropped from a height h . It hits the ground with a certain momentum ' p '. If the same stone is dropped from a different height h' such that percentage change in momentum is 41.4 % , then the height from which the stone is dropped is h ^ =x ~h , where x is:

Step-by-step solution

The momentum p of a stone dropped from height h is given by p = mv , where v = 2gh . Thus, p = m 2gh , which implies p h . Let p be the initial momentum and p' be the final momentum. The percentage change in momentum is given as 41.4 % . p' - p p = 0.414 p' p = 1 + 0.414 = 1.414 Since p h , we have p' p = h' h . Substituting the values, h' h = 1.414 . Squaring both sides, h' h = (1.414)^2 . Since 1.414 2 , we have h' h = ( 2 )^2 = 2 . Therefore, h' = 2h , which gives x = 2 . Answer: 2