Question

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. \( 2\sqrt{5} \, km \)
• B. \( 4 \, km \)
• C. \( 7.2 \, km \)
• D. \( 2\sqrt{2} \, km \)

Hint:

Remember to treat horizontal and vertical motion independently in projectile problems. The displacement is the straight-line distance between the initial and final points.

Answer 1

The Correct Option is D

Step 1: Convert the initial velocity of the object to m/s.
The initial horizontal velocity of the object is \( u_x = 100 \, m/s \).
Step 2: Calculate the horizontal distance travelled by the object.
The horizontal distance \( x \) travelled by the object is: \[ x = u_x \times t = 100 \, m/s \times 20 \, s = 2000 \, m = 2 \, km \]
Step 3: Calculate the vertical distance travelled by the object.
The vertical distance \( y \) travelled by the object is \( 2 \, km \).
Step 4: Calculate the displacement of the point O from the release point.
The horizontal displacement is \( x = 2 \, km \) and the vertical displacement is \( y = 2 \, km \) downwards. The magnitude of the displacement \( |\vec{s}| \) is: \[ |\vec{s}| = \sqrt{x^2 + y^2} = \sqrt{(2 \, km)^2 + (2 \, km)^2} = \sqrt{4 + 4} \, km = \sqrt{8} \, km = 2\sqrt{2} \, km \] The magnitude of the displacement is \( 2\sqrt{2} \, km \).