Engineering Mechanics

Chapter 27 : Collision of Elastic Bodies „„„„„ 567

and velocity with which the ball rebounds,

vgh g==×= 2 1 2 0.81 0.9 2 gm/s ...(ii)

We also know that the velocity with which the ball rebounds (v)

0.9 2ge g= 2

∴ e = 0.9 Ans.

(b) Expected height after the second bounce

Let h 2 = Expected height after the second bounce.

We know that *velocity, with which the ball impinges second time,

ug=0.9 2 m/s

and velocity, with which the ball rebounds,

vgh= (^22)
We also know that the velocity with which the ball rebounds second time (v)
2 gh 2 == ×eu 0.9 0.9 2 g=0.81 2g
∴ 2 gh 2 = (0.81)^2 2 g = 0.656 × 2g
or h 2 = 0.656 m Ans.
Example 27.10. From what height, must a heavy elastic ball be dropped on a floor, so that
after rebounding thrice it will reach a height of 16 metres? Take e = (0.5)1/3.
Solution. The system of rebounding is shown in Fig. 27.5
Fig. 27.5.
Let h 0 = Height from which the ball is dropped,
h 1 = Height after first rebound, and
h 2 = Height after second rebound

• The velocity, with which the ball impinges second time is the same with which the ball rebounded first
  time.