A bomb is dropped from an aeroplane flying horizontally with a velocity $469{\text{ms}}^{-1}$ at an altitude of $980\text{m}$. The bomb will hit the ground after a time

$(g=9.8{\text{m/s}}^2)$

The correct option is D $10\sqrt{2}\text{s}$
Height of the aeroplane from the ground,

$h=980\text{m}$

Using second equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

Taking downward as positive.

$h=u_{y}t+\frac{1}{2}g{t}^2$

$u_y=0($as bomb is dropped$)$

$\Rightarrow t=\sqrt{\frac{2h}{g}}$

$\therefore t=\sqrt{\frac{2\times 980}{9.8}}\text{s}=10\sqrt{2}\text{s}$

Hence, option $\left(\text{D}\right)$ is the correct answer.