A helicopter flying horizontally with a speed of $360 \mathrm{~km} / \mathrm{h}$ at an altitude of 2 km , drops an object at an instant. The object hits the ground at a point O , 20 s after it is dropped. Displacement of ' O ' from the position of helicopter where the object was released is :
(use acceleration due to gravity $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ and neglect air resistance)

• A

  $2 \sqrt{5} \mathrm{~km}$
• B

  4 km
• C

  7.2 km
• Correct answer
  D

  $2 \sqrt{2} \mathrm{~km}$

Solution

$\begin{aligned} & \mathrm{u}=360 \times \frac{5}{18}=100 \mathrm{~m} / \mathrm{s} \\ & \mathrm{x}=\mathrm{u} \times \mathrm{t}=2 \times 10^3 \mathrm{~m} \\ & \mathrm{t}=\sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}} \Rightarrow \mathrm{H}=\frac{\mathrm{t}^2 \mathrm{~g}}{2} \\ & \mathrm{H}=\frac{400 \times 10}{2} \\ & \mathrm{H}=2000 \mathrm{~m} \\ & \mathrm{D}=\sqrt{\mathrm{x}^2+\mathrm{H}^2} \\ & \mathrm{D}=2 \sqrt{2} \mathrm{~km}\end{aligned}$

Hence, the correct answer is option: 'D'.