Question

Two towers $A$ and $B$. each of height $20m$ are situated a distance $200m$ apart. A body thrown horizontally from the top of the tower $A$ with a velocity $20m{s}^{-1}$ towards the tower $B$ hits the ground at point $P$ and another body thrown horizontally from the top of tower $B$ with a velocity $30m{s}^{-1}$ towards the tower $A$ hits the ground at point $Q$. If a car starting from rest from $P$ reaches $Q$ in $10$ seconds, the acceleration of the car is
(Acceleration due to gravity = $10m{s}^{-2}$)

• A. $1m{s}^{-2}$
• B. $2m{s}^{-2}$
• C. $3m{s}^{-2}$
• D. $4m{s}^{-2}$

Answer 1

Answer: (B)

Given, height of both towers is same,
$h_1=h_2=h$

Time of flight, $t=\sqrt{\frac{2h}{g}}$ will be same
$t=\sqrt{\frac{2\times 20}{10}}=\sqrt{4}=2$
$\Rightarrow$ Displacement in horizontal direction from tower $A$ to point $P=u_{A}t$
$=20\times 2=40m$
$\Rightarrow$ Displacement in horizontal direction from tower $B$ to point $Q=u_{B}t$
$=30\times 2=60m$
So, distance between point $P$ and $Q$
$=200-(40+60)$
$=100m$
Given, distance between $P$ and $Q$ is covered by car in $10s$, so using
$s=ut+\frac{1}{2}a{t}^2$
$100=0\times 10+\frac{1}{2}a(10{)}^2$
$[\because u=0$, as car starts from rest $]$
$a=2$
Acceleration, $a=2m/s^2$