Question

A small boy is throwing a ball towards a wall 6m in front of him. He releases the ball at a height of 1/4m from 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 before the boy did the ball bounce from the ground?
A. 1.5m
B. 2.5m
C. 3.5m
D. 4.5m

Answer 1

Hint: Use kinematics equation. In this question, the mostly used formula is the kinematic equation. Use time of flight formula. Also use the formula of speed. Apply the concept of perfectly elastic collision.

Step by step solution:
Diagram:

Let’s assume that boy throws the ball at a horizontal speed of u and vertical speed u.
Let the distance be h where the ball bounces off the ground.
Acceleration due to gravity is given by $g=10m/{{s}^{2}}$
We know that speed is equal to distance per unit time.
Mathematically,
$speed=\dfrac{dis\tan ce}{time}$
Now time taken by the ball to hit the wall is given by,
$t=\dfrac{6}{u}$
In this time, $\Delta y=3-1.4m$=106m
Now use kinematics equation,
$\dfrac{d}{u}$$1.6=6*\dfrac{v}{u}-5{{(\dfrac{6}{u})}^{2}}$ --------(1)

Given that collision is elastic, in this situation horizontal speed changes its speed and vertical speed remain the same.

Now time taken by the ball to reach height of 1.4m is$\dfrac{12-2d}{u}$

Use time of flight formula,

$\dfrac{v}{5}=\dfrac{12-2d}{u}$ -------(2)

Time taken to reach ground is $\dfrac{d}{u}$ and vertical speed of the ball is –v for height 1.4m

Use kinematics equation in the direction of y,

$\begin{align}
  & -1.4=-v\dfrac{d}{u}-5{{(\dfrac{d}{u})}^{2}}----(3) \\
 & substitute(2)in(1)and(3) \\
 & we have, \\
 & 1.6=\dfrac{6(60-10d)-180}{{{u}^{2}}} \\
 & 1.6{{u}^{2}}=180-60d-----(4) \\
 & 8.4{{u}^{2}}=(1.6{{u}^{2}}+180)d+30{{d}^{2}}----(5) \\
 & substitute(4)in(5) \\
 & 30{{d}^{2}}-675d+945=0 \\
 & d=1.5m \\
\end{align}$
Since the value of d=21 m won’t be possible.

Note: Notice that, in the diagram there are two bounces. One on the wall and other on the ground. 6m is not a height of wall but a distance between boy and wall. You may get confused here. In this kind of question, concentration and focus are important along with a good concept. Students may get confused between kinematic equations that which kinematic equation should have to use. Do not get confused between spend and velocity as well as distance displacement.