A helicopter flies horizontally at 360 km/h at an altitude of 2 km and drops an object. The object reaches the ground after 20 s. Neglect air resistance and take g = 10 m/s². The displacement of the impact point from the release point is:
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Question

A helicopter flies horizontally at 360 km/h at an altitude of 2 km and drops an object. The object reaches the ground after 20 s. Neglect air resistance and take g = 10 m/s². The displacement of the impact point from the release point is:

Options

G1 Physics · Accepted Answer

Horizontal speed of the object at release equals the helicopter speed: u x u_x u x ​ = 360 km/h = 360×(5/18) = 100 m/s. In 20 s, horizontal displacement is x = u x t u_xt u x ​ t = 100×20 = 2000 m = 2 km. Vertical displacement is the altitude: H = 2 km. Therefore the straight-line displacement from release point to impact point is D = √(x²+H²) = √(2²+2²) km = 2√2 km. Correct option: D.