JEE Physics 2025 Question with Solution

Given: The helicopter moves horizontally with speed $360 \, \text{km/h}$, the object reaches the ground after $20 \, \text{s}$, and $g = 10 \, \text{m/s}^2$.

Find: The displacement of point O from the release point of the object.

Treat horizontal and vertical motions independently.

First convert the horizontal speed:

$$360 \, \text{km/h} = \frac{360}{3.6} \, \text{m/s} = 100 \, \text{m/s}$$

Horizontal distance covered in $20 \, \text{s}$:

$$x = u_x t = 100 \times 20 = 2000 \, \text{m} = 2 \, \text{km}$$

Vertical distance covered in free fall:

$$y = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times (20)^2 = 2000 \, \text{m} = 2 \, \text{km}$$

Now the displacement is the straight-line distance between the release point and point O:

$$d = \sqrt{x^2 + y^2} = \sqrt{(2 \, \text{km})^2 + (2 \, \text{km})^2}$$ $$d = \sqrt{4 + 4} \, \text{km} = \sqrt{8} \, \text{km} = 2\sqrt{2} \, \text{km}$$

Therefore, the displacement is $2\sqrt{2} \, \text{km}$, so the correct option is D.

Given: Initial horizontal velocity is equal to the helicopter speed, the initial vertical velocity is zero, time of flight is $20 \, \text{s}$, and $g = 10 \, \text{m/s}^2$.

Find: The magnitude of displacement from the release point to the impact point.

Since the object is dropped, not projected upward or downward, its initial vertical velocity is zero. The horizontal component remains constant throughout the motion.

1. Convert speed of the helicopter:

$$u_x = 360 \, \text{km/h} = 100 \, \text{m/s}$$

2. Horizontal displacement:

$$x = v_x t = 100 \, \text{m/s} \times 20 \, \text{s} = 2000 \, \text{m}$$

3. Vertical displacement:

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

4. Magnitude of resultant displacement:

$$|\vec{s}| = \sqrt{x^2 + y^2} = \sqrt{(2000)^2 + (2000)^2} \, \text{m}$$ $$|\vec{s}| = 2000\sqrt{2} \, \text{m} = 2\sqrt{2} \, \text{km}$$

Thus, the magnitude of displacement is $2\sqrt{2} \, \text{km}$. Hence, the correct option is D.