A Stone falls from a high of 10m on hard horizontal surface. if 25% of energy is lose due to impact.how high will it rise?

Asked by kuriakose7600 | 24th Nov, 2015, 06:50: PM

Expert Answer:

Given that the stone falls from height, h = 10m. 
P.E at this height = mgh
On reaching the ground, K.E = mgh
Since, the stone loses 25% of its energy due to impact, K.E due to one impact will be 0.75 mgh.
Let v1 be the initial upward velocity of the stone after impact, we get
begin mathsize 12px style 1 half mv subscript 1 superscript 2 space equals space 0.75 space mgh Or space 1 half mv subscript 1 superscript 2 space equals space 3 over 4 space mgh Or space straight v subscript 1 superscript 2 space equals space 1.5 space gh space minus negative negative negative negative space left parenthesis Equation space 1 right parenthesis Thus comma space the space height space straight h subscript 1 space to space which space the space stone space will space rise space is space straight h subscript 1 space equals space fraction numerator straight v subscript 1 superscript 2 over denominator 2 straight g end fraction space space space space space equals space fraction numerator 1.5 space gh over denominator 2 straight g end fraction space space minus negative negative negative negative space left parenthesis From space equation space 1 right parenthesis space space space space space space space equals space 0.75 space straight h straight h subscript 1 space equals space 0.75 space cross times space 10 space space minus negative negative negative negative space left parenthesis space straight h space equals space 10 space straight m right parenthesis space straight h subscript 1 space equals space 7.5 space straight m end style
Thus, the stone will rise to height of 7.5 m after the first impact.

Answered by Yashvanti Jain | 25th Nov, 2015, 11:49: AM