Question

5. A highly elastic ball is released from rest at a distance h above the ground and bounces up and
down. With each bounce a fraction f = 1/10 of its KE just before the bounce is lost. The time for
which the ball will continue to bounce is nearly (take g = 10m/s2, h = 5m)
1) 1 sec
2) 38 sec
3) 84.3 sec
4) 1.86 sec​

Answer 1

Answer:

fdngahsheje I am I supposed I would like a call to the intended addressee email report rue la the right away so much help full for a good idea for you and I received the first to do this ritoriitoiiititititi

Answer 2

Answer :

38 seconds

Explanation :

$t = \sqrt{\frac{2h}{g} }( \frac{1 + e}{ 1 - e\\} )$  

$K_i$ = $= \frac{1}{2} mv^{2}$

After first collision

$K_f = \frac{9}{10} (\frac{1}{2}mv^{2} )$

Velocity changes e times of previous velocity every time

$V_2 = e^{n} V_1$

$\sqrt{\frac{9}{10}} V = e^{n}V$    $=>$    $e = \sqrt[]{\frac{9}{10} }$

Substituting e in the equation of time, we get :

$t = 38 sec$