A helicopter is flying horizontally at an altitude of $2\text{km}$ with a speed of $100{\text{ms}}^{-1}$. A packet is dropped from it. The horizontal distance between the point where the packet is dropped and the point where it hits the ground is $(g=10{\text{ms}}^{-2})$

The correct option is A $2\text{km}$
Height of the helicopter from the ground $=2000\text{m}$

Using second equation of motion,
$h=u_{y}t+\frac{1}{2}g{t}^2$
Since $u_y=0$, $t=\sqrt{\frac{2h}{g}}$
$\Rightarrow t=\sqrt{\frac{2\times 2000}{10}}\text{s}=\sqrt{400}\text{s}=20\text{s}$
$\therefore$ Horizontal distance travelled by the packet
$x=ut=100{\text{ms}}^{-1}\times 20\text{s}=2000\text{m}=2\text{km}$