A ball is dropped from height h on a horizontal floor. If it loses 60% of its energy on hitting the floor then height up-to which it will rise after first renounce is? h/5. 2h/5. 3h/5. 4h/5

Hi Tasneem dear, energy = 1/2 * m * v^2
v^2 = 2 g h ----(1)
1/2 m v^2 = 100%
Now with a new speed bouncing speed u, kinetic energy = 1/2 * m * u^2 = 40%
So dividing u^2 / v^2 = 40/100 = 2/5
Again u^2 = 2 g x ( here x is the height to be found)
Dividing u^2 /v^2 = x/h
==> x/h = 2/5
OR x = 2/5 * h 
Hence second option
 

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