Question

A particle is projected with a speed $u$ at an angle $\theta$ with horizontal from point $A$. It strikes elastically with a vertical wall at height $h/2$. It rebounds and reaches maximum height $h$ and falls back on the ground at point $B$ as shown in Figure. Distances from $A$ to wall and from wall to $B$ are $x_1$ and $x_2$, and time to cover is $t_1$ and $t_2$ respectively. Match the values in column I with the expressions in column II.


$$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{i.}\sqrt{2}& \text{a.}\frac{x_2-x_1}{x_2+x_1}\text{or}\frac{x_2+x_1}{x_2-x_1}\\ \text{ii.}\frac{1}{\sqrt{2}}& \text{b.}\frac{t_2-t_1}{t_2+t_1}\text{or}\frac{t_2+t_1}{t_2-t_1}\\ \text{iii.}1& \text{c.}\frac{u\sin \theta }{g(t_2+t_1)}\\ \text{iv.}\frac{1}{2}& \text{d.}\frac{u\cos \theta (t_1+t_2)}{x_1+x_2}\end{array}$$

• A. $\text{(i) a, b (ii) a, b (iii) d (iv) c}$
• B. $\text{(i) a (ii) b (iii) d (iv) c}$
• C. $\text{(i) a, b (ii) a, b (iii) d (iv) c}$
• D. $\text{(i) a (ii) a, c (iii) d (iv) b}$

Answer 1

The correct option is A $\text{(i) a, b (ii) a, b (iii) d (iv) c}$

Distance covered $x_1$ and $x_2$ will be
$\begin{array}{}ucos\theta t_1=x_1& \text{(1)}\end{array}$
$\begin{array}{}ucos\theta t_2=x_2& \text{(2)}\end{array}$
Dividing $\left(1\right)$ and $\left(2\right)$
$\begin{array}{}\frac{t_1}{t_2}=\frac{x_1}{x_2}& \text{(3)}\end{array}$
At maximum height, velocity has only horizontal component.
Hence the vertical distance covered by the body will be proportional to square of time taken to cover that distance.
$h\propto t^2$
If $T$ is the time taken for the whole projectile motion, than time taken to rach maximum height $h$ will be $T/2$.
If $t_1$ is the time taken to cover $x_1$, then time taken to cover $h/2$ is $(\frac{T}{2}-t_1)$
Since $h\propto t^2$
$\frac{h}{h/2}=\frac{T/2}{\frac{T}{2}-1}$
$\begin{array}{}\Rightarrow (\frac{T}{T-2t_1}{)}^2=2& \text{(4)}\end{array}$
Now total time of flight, $T=t_1+t_2$
Using this in $\left(3\right)$,
$\begin{array}{}\frac{t_2+t_1}{t_2-t_1}=\sqrt{2}& \text{(5)}\end{array}$
From $\left(3\right)$ we can write,
$\frac{t_2}{t_1}=\frac{x_2}{x_1}$
$\frac{t_2+t_1}{t_2-t_1}=\frac{x_2+x_1}{x_2-x_1}$
$\frac{x_2+x_1}{x_2-x_1}=\sqrt{2}$

$\begin{array}{}\frac{usin\theta }{g(t_1+t_2)}=\frac{usin\theta }{Tg}& \text{(6)}\end{array}$
Since the time of flight for projectile, $T=\frac{2usin\theta }{g}$
Using this in $\left(6\right)$,
$\frac{usin\theta }{g(t_1+t_2)}=\frac{usin\theta }{\frac{2usin\theta }{g}g}=\frac{1}{2}$
Since $t_1+t_2=T$ and $x_1+x_2=R$
$\frac{ucos\theta (t_1+t_2)}{x_1+x_2}=\frac{ucos\theta T}{R}=\frac{ucos\theta T}{ucos\theta T}=1$