(^568) „„„„„ A Textbook of Engineering Mechanics
We know that the velocity with which the ball impinges on the floor,
ugh= 2 0 ...(i)
and the velocity with which the ball rebounds first time
vgh= 2 1 ...(ii)
Similarly, velocity with which the ball impinges after first rebound,
uv gh 11 == 2 ...(iii)
and velocity with which the ball rebound second time
vgh 12 = 2
Similarly, velocity with which the ball impinges after second rebound,
uv 21 == 2 gh 2 ...(iv)
and velocity with which the ball rebounds third time
ug 3 =×=21642m/sg ...(v)
We know that during first impact,
v 1 = eu 1 ...(vi)
or 2(0.5)2gh 10 = 1/ 3 gh .. .(vii)
Similarly, during second impact,
2
(^1) 1/ 3
2
2
(0.5)
gh
gh = ...(viii)
1/ 3
2(0.5)2gh 21 = gh ...(ix)
and during third impact
1/3
216(0.5)ggh×= × (^22)
or (^2) 1/ 3 1/ 3
21642
2
(0.5) (0.5)
g g
gh
×

Substituting the value of 2 gh 2 in equation (viii),
(^1) 1/3 1/ 3 2/3
42 4(2)
2
(0.5) (0.5) (0.5)
gg
gh ==
×
Now substituting the value of 2 gh 1 in equation (vii)
1/ 3
2/3^0
4(2)
(0.5) 2
(0.5)
g
= gh
0
4
8
(0.5)
h ==
h 0 = 64 m Ans.