Physics Q29 – JEE-MAIN 2025

Home Physics Motion in a PlaneQuestion #312

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)

Clue 1 of 3: Projectile Motion Components

Remember that horizontal and vertical motions are independent in projectile motion, assuming no air resistance.

🥷

Ninja Strategy•Pythagorean Estimate & Unit Check

Recognize that the horizontal and vertical displacements are $2 km$ each, so the total displacement must be $2 \sqrt{2} km \approx 2.8 km$ . Also, ensure unit consistency to avoid large errors like $7.2 km$ .

🧠 Full Solution Path

Step 1: Convert Units and Identify Components✦ Active

First, convert the given speed from $k m / h$ to $m / s$ and altitude from $k m$ to $m$ for consistency with $g$ and time in seconds. Identify the initial horizontal velocity and the vertical displacement.

u_{x} = 360 km/h = 360 \times \frac{1000}{3600} m/s = 100 m/s

h = 2 km = 2000 m

t = 20 s

g = 10 {m/s}^{2}

💡 Teacher's Secret Hint

Ensure all units are consistent (SI units are usually best) before performing calculations.

Step 2: Calculate Horizontal Displacement○ Expand

Since air resistance is neglected, the horizontal velocity remains constant. Calculate the horizontal distance covered by the object during the $20 s$ it takes to hit the ground.

x = u_{x} \times t

x = 100 m/s \times 20 s = 2000 m = 2 km

💡 Teacher's Secret Hint

The time of flight is determined by the vertical motion, but it applies to both horizontal and vertical components.

Step 3: Calculate Total Displacement○ Expand

The total displacement is the straight-line distance from the point of release to the point O on the ground. This can be found using the Pythagorean theorem, as the horizontal and vertical displacements form a right-angled triangle.

D = \sqrt{x^{2} + h^{2}}

D = \sqrt{(2000 m)^{2} + (2000 m)^{2}}

D = \sqrt{2 \times (2000 m)^{2}} = 2000 \sqrt{2} m

D = 2 \sqrt{2} km

💡 Teacher's Secret Hint

Displacement is a vector quantity, and its magnitude is the shortest distance between the initial and final points.

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Physics Question 29 – JEE-MAIN 2025

Clue 1 of 3: Projectile Motion Components

🧠 Full Solution Path