A ball falling freely from a height of 4.9 m/s, hits a horizontal surface. If $e=\frac{3}{4},$ , then the ball will hit the surface, second time after

Velocity on hitting the surface$=\sqrt{2\times 9.8\times 49}=9.8 m/s$ 

Velocity after first bounce

     $v=\frac{3}{4}\times 9.8$

Time taken from first bounce to the second bounce$e=\frac{2v}{g}$$=2\times \frac{3}{4}\times 9.8\times \frac{1}{9.8}=1.5\mathrm{s}$