Question

A cart is moving horizontally along a staight line with a constant speed $30\text{m/s}$. A bullet is to be fired vertically upwards from the moving cart in such a way that it returns to the cart at the same point from where it was fired, after the cart has moved $80\text{m}$. At what speed (relative to the cart), the bullet must be fired ? Take $g=10{\text{m/s}}^2$.

• A. $10\text{m/s}$
• B. $10\sqrt{8}\text{m/s}$
• C. $\frac{40}{3}\text{m/s}$
• D. $\frac{16}{\sqrt{3}}\text{m/s}$

Answer 1

The correct option is C $\frac{40}{3}\text{m/s}$

velocity of cart w.r.t ground is $\overrightarrow{v_{C,G}}=30\text{m/s}\hat{i}$
$\Rightarrow$ Let velocity of bullet with respect to cart is,
$\overrightarrow{v_{B,C}}=u_y\text{m/s}\hat{j}$
Hence, velocity of bullet with respect to ground $\left(\overrightarrow{v_{B,G}}\right)$ is,
$\overrightarrow{v_{B,G}}=\overrightarrow{v_{B,C}}+\overrightarrow{v_{C,G}}$
$\Rightarrow \overrightarrow{v_{B,G}}=(u_y\hat{j}+30\hat{i})\text{m/s}$
$\overrightarrow{a_{B,G}}=-g\hat{j}=-10{\text{m/s}}^2\hat{j}$
$\because$ Fired bullet has velocity component in $x\&y$ directions w.r.t ground, and its acceleration is $g$ downwards, hence bullet will follow projectile motion.

$\Rightarrow$Time of flight for bullet will be,
$T=\frac{2u_y}{g}...\left(i\right)$
Horizontal distance covered till it is in air,
$R=(v_{B,G}{)}_x\times T...(ii)$
since bullet returns to cart when it has travelled $80\text{m}$
$R=80\text{m}...\left(iii\right)$
From Eq $\left(i\right),\left(ii\right),\left(iii\right)$,
$80=(v_{B,G}{)}_x\times \frac{2u_y}{g}$
$80=30\times \frac{2u_y}{10}$
$\therefore u_y=\frac{40}{3}\text{m/s}$