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 determine the displacement of point O from the position of the helicopter where the object was released, let's break down the problem:

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

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

• Given:
  • Acceleration due to gravity, \(g = 10 \, \text{m/s}^2\)
  • Time, \(t = 20 \, \text{s}\)
• Substituting the given 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: The horizontal speed of the helicopter is converted to meters per second: \(\frac{360 \, \text{km/h}}{3.6} = 100 \, \text{m/s}\)
2. Since the horizontal velocity is constant, use: \(d_{\text{horizontal}} = v_{\text{horizontal}} \times t\)
3. Substituting the given 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 diagonal of the right triangle formed by the vertical and horizontal distances.
   • Using Pythagorean Theorem:

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

• Calculating further:

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

Thus, the displacement of 'O' from the position of the helicopter where the object was released is \(2\sqrt{2} \, \text{km}\), which matches option \(2\sqrt{2} \, \text{km}\).

Answer 2

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 \( 2\sqrt{2} \, km \).