Question

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 $(g={\text{10 ms}}^{-2})$

• A. $2\text{km}$
• B. $0.2\text{km}$
• C. $20\text{km}$
• D. $4\text{km}$

Answer 1

The correct option is A $2\text{km}$

Height of the helicopter from the ground = $2000\text{m}$

Using second equation of motion,
$h=u_{y}t+\frac{1}{2}g{t}^2$
$u_y=0$ (initial velocity is zero)
$\Rightarrow t=\sqrt{\frac{2h}{g}}$
$\Rightarrow t=\sqrt{\frac{2\times 2000}{10}}\text{s}=\sqrt{400}\text{s}=20\text{s}$

Therefore, horizontal distance travelled by packet
$x=u_{x}t=100{\text{ms}}^{-1}\times 20\text{s}=2000\text{m}=2\text{km}$

Alternative solution:

Horizontal distance covered by a body falling from height $h$ with initial horizontal velocity $u$
$x=u\sqrt{\frac{2h}{g}}=100\times \sqrt{\frac{2\times 2000}{10}}=2000\text{m}=2\text{km}$