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 :

 ❓ 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 seconds after it is dropped. Find the displacement of point O from the position of the helicopter where the object was released.

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🖼️ Question Image

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✍️ Short Solution

Let’s calculate step-by-step:

Step 1: Horizontal Distance (Range)

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

Step 2: Vertical Motion Check

Vertical displacement under gravity:

$$y=\frac{1}{2}g{t}^2=\frac{1}{2}\times 9.8\times {20}^2=1960\text{m}$$

which is approximately equal to the helicopter’s altitude (2000 m).
✅ Hence, the time given (20 s) is consistent.

Step 3: Resultant Displacement

The object moves 2000 m horizontally and 2000 m vertically downward,
so total displacement:

$$s = \sqrt{x^2 + y^2} = \sqrt{(2000)^2 + (2000)^2} = 2000\sqrt{2}\,\text{m}$$
$$\boxed{{\displaystyle s=2.83\text{ km (approximately)}}}$$

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🧮 Image Solution

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✅ Conclusion & Video Solution

✅ Final Answer:

$$\text{Displacement}=2000\sqrt{2}\text{ m}\approx 2.83\text{ km}$$

Concept Used:

• Uniform horizontal velocity → $x = u_x t$
  

• Vertical motion under gravity → $y = \frac{1}{2}gt^2$
  

• Displacement → $s = \sqrt{x^2 + y^2}$