The correct option is A √2hg(1+e1−e)
When the ball is dropped (u=0) from height h, time taken to reach the floor will be,
t0=√2hg
Its speed while striking floor is
v0=√2gh
After 1stcollision with floor, its rebound speed will be:
v1=ev0
Total time (t1) between 1st & 2nd collision is equal to time of flight of ball, after 1st collision.
∴t1=2v1g
Similarly, total time (t2) between 2nd & 3rd collision is equal to time of flight of ball, after 2nd collision.
t2=2v2g, where v2=ev1
(since ball will hit the ground at same velocity with which its starts)
Now, v2=ev1=e(ev0)=e2v0
Similarly, v3=e3v0
Thus, total time after which ball will stop rebounding is given by,
ΔT=t0+t1+t2+t3+.....∞
ΔT=t0+2v1g+2v2g+2v3g+....∞
ΔT=t0+2ev0g+2e2v0g+2e3v0g+....∞
Putting value of v0 in above equation,
ΔT=√2hg+2ev0g(1+e+e2+e3+.....∞)
ΔT=√2hg+2e√2hg(1+e+e2+e3+.....∞)
ΔT=√2hg[1+2e(1+e+e2+e3+.....∞)] ....(i)
Applying formula for sum of infinite GP, where a=1, r=e
∑=a1−r=11−e ...(ii)
From Eq (i) & (ii):
ΔT=√2hg[1+2e(11−e)]
∴ΔT=√2hg(1+e1−e)