A bomb is dropped on an enemy post by an aeroplane flying horizontally with a velocity of $60{\text{kmh}}^{-1}$ and at a height of $490\text{m}$. At the time of dropping the bomb, how far the aeroplane should be from the enemy post so that the bomb may directly hit the target?

The correct option is B $\frac{500}{3}\text{m}$
Given, velocity $\left(v\right)=60\text{km/hr}.$
When the bomb is being dropped, it will acquire the velocity of the plane.

Hence the initial horizontal velocity of the bomb will be, $u_x=60\text{km/hr}$ and time taken for the bomb to reach the ground will be (taking downward direction as positive), $s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$, here $s_y=h$ will be the displacement in vertical direction, $u_y=0$, as the initial velocity in vertical direction $0$ and $a_y=g$
we get $h=\frac{1}{2}g{t}^2,\Rightarrow t=\sqrt{\frac{2h}{g}}$

height of projectile $h=490\text{m}$

Range of projectile $\left(R\right)=u_x\times t=u_x\sqrt{\frac{2h}{g}}$

$=60\times \frac{5}{18}\times \sqrt{\frac{2\times 490}{9.8}}$, as $(1\text{km/hr}=\frac{5}{18}\text{m/s})$

$=\frac{300\times 10}{18}=\frac{500}{3}\text{m}$

Hence option B is the correct answer.