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

To calculate the displacement of point O from the helicopter's release position, we follow these steps:

1. The object is released from a horizontally moving helicopter. It possesses an initial horizontal velocity but no initial vertical velocity, as it is dropped, not thrown, from rest relative to the helicopter.
2. Determine the time of flight: The object takes \(20 \, \text{s}\) to reach the ground.
3. Calculate the vertical distance using the free-fall formula:

\(s = \frac{1}{2} g t^2\)

• Given:
  • Acceleration due to gravity, \(g = 10 \, \text{m/s}^2\)
  • Time, \(t = 20 \, \text{s}\)
• Substitute values:

\(s = \frac{1}{2} \times 10 \, \text{m/s}^2 \times (20 \, \text{s})^2 = 2000 \, \text{m} = 2 \, \text{km}\)

1. Calculate the horizontal distance: Convert the helicopter's horizontal speed to meters per second: \(\frac{360 \, \text{km/h}}{3.6} = 100 \, \text{m/s}\)
2. Since horizontal velocity is constant, use the formula: \(d_{\text{horizontal}} = v_{\text{horizontal}} \times t\)
3. Substitute values:

\(d_{\text{horizontal}} = 100 \, \text{m/s} \times 20 \, \text{s} = 2000 \, \text{m} = 2 \, \text{km}\)

1. Determine the total displacement: The displacement is the hypotenuse of the right triangle formed by the vertical and horizontal distances.
   • Apply the Pythagorean Theorem:

\(d = \sqrt{(d_{\text{horizontal}})^2 + (s)^2} = \sqrt{(2 \, \text{km})^2 + (2 \, \text{km})^2}\)

• Perform calculation:

\(d = \sqrt{4 + 4} = \sqrt{8} \, \text{km} = 2\sqrt{2} \, \text{km}\)

Therefore, the displacement of 'O' from the helicopter's release position is \(2\sqrt{2} \, \text{km}\), matching option \(2\sqrt{2} \, \text{km}\).