Question

A small boy is throwing a ball towards a wall 6 m in front of him. He releases the ball at a height of 1.4 mfrom the ground. The ball bounces from the wall at a height of 3m, rebounds from the ground and reaches the boy's hand exactly at the point of release. Assuming the two bounces (one from the wall and the other from the ground) to be perfectly elastic, how far ahead of the boy did the ball bounce from the ground? ( plze answer in detail)​

Answer 1

Answer:

Let us assume the boy throws the ball at a Horizontal speed $$u$$ and a vertical speed $$v$$.

Let the distance where the ball bounces off the ground be $$d$$.

$$g = 10 m/s^2$$

Time taken by the ball to hit the wall = $$\frac{6}{u}$$

In this time, $$\Delta y = 3 - 1.4 = 1.6m$$

Using Kinematics relations, $$1.6 = 6\frac{v}{u} - 5 (\frac{6}{u})^2 $$ ---------(1)

As the collision is elastic at the wall, horizontal speed changes its sign, vertical speed remains the same.

Time taken by the ball to reach a height of $$1.4 m$$ is $$\frac{12-2d}{u}$$

From time of flight formula, $$\frac{v}{5} = \frac{12-2d}{u}$$ -------------------(2)

At the height of $$1.4m$$ the vertical speed of the ball is $$-v$$ and time taken to reach the ground is $$\frac{d}{u}$$

Applying kinematics relations in the $$y$$ direction,

$$-1.4 = -v\frac{d}{u} - 5(\frac{d}{u})^2$$ -------------------(3)

Substituting (2) in (1) and (3), we have

$$1.6 = \frac{6(60-10d)-180}{u^2} \Rightarrow 1.6u^2 = 180-60d$$ ----------(4)

and $$8.4u^2 = (1.6u^2+180)d+30d^2$$ ------------(5)

Substituting (4) in (5)

$$30d^2 - 675d +945=0 \Rightarrow d = 1.5m$$ $$\textrm{or}$$ $$d = 21m$$

As $$d=21m$$ is not possible, $$d = 1.5m$