A bomber plane moves horizontally with a speed of 500 m/s and a bomb released from it, strikes the ground in 10 sec. Angle at which it strikes the ground will be $(g=10m/s^2)$

The correct option is A

Horizontal component of velocity $V_x$ = 500m/s
and vertical components of velocity while striking the ground.

$V_y$ = $0+10\times 10$ = $100m/s$
$\therefore$ Angle with which it strikes the ground.
$\theta$ = $ta{n}^{-1}$$\left(\frac{v_y}{v_x}\right)$ = $ta{n}^{-1}$$\left(\frac{100}{500}\right)$ = $ta{n}^{-1}$$\left(\frac{1}{5}\right)$