JEE MAIN - Physics (2025 - 7th April Evening Shift - No. 3)
A helicopter flying horizontally with a speed of 360 km/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 g = 10 m/s2 and neglect air resistance)
Explanation
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$$\begin{aligned} & u=360 \times \frac{5}{18}=100 \mathrm{~m} / \mathrm{s} \\ & \mathrm{x}=\mathrm{u} \times \mathrm{t}=2 \times 10^3 \mathrm{~m} \\ & \mathrm{t}=\sqrt{\frac{2 \mathrm{H}}{\mathrm{~g}}} \Rightarrow \mathrm{H}=\frac{\mathrm{t}^2 \mathrm{~g}}{2} \\ & \mathrm{H}=\frac{400 \times 10}{2} \\ & \mathrm{H}=2000 \mathrm{~m} \\ & \mathrm{D}=\sqrt{\mathrm{x}^2+\mathrm{H}^2} \\ & \mathrm{D}=2 \sqrt{2} \mathrm{~km} \end{aligned}$$
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