JEE Main 2025 Apr 7 Shift 2, Physics Q18: Motion In Two And Three Dimension

JEE Main2025Apr 7, Shift 2Physics

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)

A
$25$ km
B
4 km
C
7.2 km
D
$22$ km

Solution

Converting the helicopter's speed: $u = 360$ km/h $= 360 \times \frac{5}{18} = 100$ m/s. The horizontal displacement in $t = 20$ s is $x = u t = 100 \times 20 = 2000$ m $= 2$ km.

The vertical distance fallen in 20 s under gravity is $y = \frac{1}{2} g t^{2} = \frac{1}{2} (10) (400) = 2000$ m $= 2$ km, which correctly matches the given altitude, confirming the setup.

The displacement from the release point to the landing point O is the straight-line distance: $D = x^{2} + y^{2} = 2^{2} + 2^{2} = 22$ km.

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JEE Main 2025 Apr 7 Shift 2, Physics Q18: Motion In Two And Three Dimension

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