A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to 81100 of the height through which it falls. Find the average speed of the ball. (Take g=10 ms−2 )

Let us say initial height is h .Velocity of ball just before hitting floor, u=2ghVelocity of ball just after hitting floor, V=euLet us say new height achieved is h'.So,V=2gh'⇒eu=2gh'⇒e2gh=2gh'Given : h'=81100h⇒e=0.9Total distance travelled by ball=S=h+2e2h+2e4h+2e6h...=h1+e21-e2=47.632 mTotal time taken = t=2hg+22e2hg+22e4hg+22e6hg=2hg1+e1-e=19 sAverage Speed , Vavg= Total distance  Total time =St ⇒Vavg=2.5 ms−1