Question

Two towers $A$ and $B$. each of height $20\, m$ are situated a distance $200\, m$ apart. A body thrown horizontally from the top of the tower $A$ with a velocity $20 \, ms^{-1}$ towards the tower $B$ hits the ground at point $P$ and another body thrown horizontally from the top of tower $B$ with a velocity $30 \, ms^{-1}$ 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 = $10 \, ms^{-2}$)

• A. $1 \, ms^{-2}$
• B. $2 \, ms^{-2}$
• C. $3 \, ms^{-2}$
• D. $4 \, ms^{-2}$

Answer 1

Answer: (B)

Given, height of both towers is same,
$h_{1}=h_{2}=h$

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