Question

A stone is projected from the ground and hits a smooth vertical wall after $1\text{s}$ and again fall back on the ground as shown in the figure. The time taken by the stone to reach the ground after the collision is $3\text{s}$. The maximum height reached by the same stone if the vertical wall is not present will be
(Take $g=10{\text{m/s}}^2$)

• A. $14\text{m}$
• B. $16\text{m}$
• C. $18\text{m}$
• D. $20\text{m}$

Answer 1

The correct option is D $20\text{m}$

Time taken by stone to reach ground is,
$T=1+3=4\text{s}$
On collision with wall, only the horizontal component of velocity changes, hence time period will remain same.
So, $T=\frac{2u\sin \theta }{g}=4\text{s}$
$\Rightarrow u\sin \theta =2g$ ............$\left(1\right)$
Now, maximum height reached in absence of wall,
$H=\frac{u^2{\sin }^2\theta }{2g}$
$\Rightarrow H=\frac{4g^2}{2g}$ $[$from $\left(1\right)]$
$\Rightarrow H=20\text{m}$

$\begin{array}{c}\text{Why this question?}\\ \text{Tips: On collision with a smooth vertical wall, only the horizontal component of velocity gets changed as impulse by wall is in the normal direction, thus the vertical component of velocity remains same. Time of flight does not change.}\end{array}$