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

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Solution

Q.
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 $m / s^{2}$ and neglect air resistance) [2025]

1 $2 \sqrt{5}$ km

2 4 km

3 7.2 km

4 $2 \sqrt{2}$ km

Ans.
(4)

$u = 360 \times \frac{5}{18} = 100 m / s$

$x = u \times t = 100 \times 20 = 2 \times 10^{3} m = 2 km$

$t = \sqrt{\frac{2 H}{g}} \Rightarrow H = \frac{t^{2} g}{2}$

$H = \frac{400 \times 10}{2} = 2000 m = 2 km$

Displacement, $D = \sqrt{x^{2} + H^{2}} = 2 \sqrt{2} km$ .

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