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)
(use acceleration due to gravity $ g = 10 \, m/s^2 $ and neglect air resistance)
Show Hint
- \( 2\sqrt{5} \, km \)
- \( 4 \, km \)
- \( 7.2 \, km \)
- \( 2\sqrt{2} \, km \)
The Correct Option is D
Approach Solution - 1
To determine the displacement of point O from the position of the helicopter where the object was released, let's break down the problem:
- 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.
- Calculate the time of flight: The object takes \(20 \, \text{s}\) to reach the ground.
- 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:
\(s = \frac{1}{2} \times 10 \, \text{m/s}^2 \times (20 \, \text{s})^2 = 2000 \, \text{m} = 2 \, \text{km}\)
- Calculate the horizontal distance: The horizontal speed of the helicopter is converted to meters per second: \(\frac{360 \, \text{km/h}}{3.6} = 100 \, \text{m/s}\)
- Since the horizontal velocity is constant, use: \(d_{\text{horizontal}} = v_{\text{horizontal}} \times t\)
- Substituting the given values:
\(d_{\text{horizontal}} = 100 \, \text{m/s} \times 20 \, \text{s} = 2000 \, \text{m} = 2 \, \text{km}\)
- 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 \(2\sqrt{2} \, \text{km}\).
Approach Solution -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 \( 2\sqrt{2} \, km \).
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