JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2025
- Medium
- MCQ
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)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(2 \sqrt{2} \mathrm{~km}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{u}=360 \times \frac{5}{18}=100 \mathrm{~m} / \mathrm{s} \\ & \mathrm{x}=\mathrm{u} \times \mathrm{t}=2 \times 10^3 \mathrm{~m} \\ & \mathrm{t}=\sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}} \Rightarrow \mathrm{H}=\frac{\mathrm{t}^2 \mathrm{~g}}{2} \\ & \mathrm{H}=\frac{400 \times 10}{2} \\ & \mathrm{H}=2000 \mathrm{~m} \\ & \mathrm{D}=\sqrt{\mathrm{x}^2+\mathrm{H}^2} \\ & \mathrm{D}=2 \sqrt{2} \mathrm{~km}\end{aligned}\)
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