An object is dropped from a helicopter flying horizontally at 360 km/h. It falls from a height of 2 km and reaches the ground in 20 seconds. What is the displacement of the package relative to the helicopter's position when it was dropped?

Question

A particle of mass $m$ is projected with a velocity $u$ making an angle of $30^\circ$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is _____.

Answer 1

An object is released horizontally from a helicopter. The object's initial horizontal velocity is the same as the helicopter's horizontal velocity. Vertical motion is governed by free fall principles. Horizontal distance is calculated as: \[ d = v_x t \] In this formula, $ d $ represents horizontal displacement, $ v_x $ is the horizontal velocity, and $ t $ is the time. The given values are $ v_x = 360 \, \text{km/h} $ and $ t = 20 \, \text{s} $. Convert $ v_x $ to meters per second: \[ v_x = 360 \times \frac{1000}{3600} = 100 \, \text{m/s} \] Therefore, the horizontal displacement is: \[ d = 100 \times 20 = 2000 \, \text{m} = 2 \, \text{km} \] The total displacement from the helicopter's perspective is $ 2\sqrt{2} \, \text{km} $.

An object is dropped from a helicopter flying horizontally at 360 km/h. It falls from a height of 2 km and reaches the ground in 20 seconds. What is the displacement of the package relative to the helicopter's position when it was dropped?

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The Correct Option is B

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