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 $1.4$ m from the ground. The ball bounces from the wall at a height of $3$ m, rebounds from the ground a n d reaches the boy’s h an d exactly a t 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?

• A. 1.5 m
• B. 2.5 m
• C. 3.5 m
• D. 4.5 m

Answer 1

Answer: (A)

Motion of the ball is as shown below.

We can combine all three portions one after other to get a perfect parabola obtained by throwing ball from a point on the ground. It is as shown below.

Now, for the standard parabolic motion, equation of trajectory can be written as
$y=x\tan \theta -\frac{g{x}^2}{2{\upsilon }^2{\cos }^2\theta }$
Substituting $R=\frac{u^{2}sin2\theta }{g}$ in above
equation, we have
$Y=X\tan \theta .\left(1-\frac{X}{R}\right)$
When $y=1.4m,X=x$ and $R=12m$, so
We have $1.4=x\tan \theta .\left(1-\frac{x}{12}\right)\cdots \left(i\right)$
Again, when $y=3m$ and $X=6+x$, so We have
$3=\left(6+x\right)\tan \theta \cdot \left(1-\frac{6+x}{12}\right)...\left(i\right)$
Dividing Eq. $\left(i\right)$ by Eq, $\left(ii\right)$, we get
$\frac{1.4}{3}=\frac{x\left(12-x\right)}{\left(6+x\right)\left(6-x\right)}$
$\Rightarrow 7\left(36-x^2\right)=15\left(12x-x^2\right)$
$\Rightarrow 2x^2-45x+63=0$
$\Rightarrow x^2-\left(\frac{3}{2}+21\right)x+\left(\frac{3}{2}\times 21\right)=0$
$\Rightarrow x^2-\frac{3}{2}x-21x+\frac{3}{2}\times 21=0$
$\Rightarrow \left(x-\frac{3}{2}\right)-21\left(x-\frac{3}{2}\right)=0$
$\Rightarrow \left(x-\frac{3}{2}\right)\left(x-21\right)=0$
So, $x=\frac{3}{2}$ or $x=21$
ASs $x$ muist be less than $6m$
So, $x=\frac{3}{2}=1.5m$