Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A balloon is at a height of $81 \text{ m}$ and is ascending upwards with a velocity of $12 \text{ m/s}$ . A body of $2 \text{ kg}$ weight is dropped from it. If $g=10 \text{ m/s}^2$ , the body will reach the surface of the earth in:

1.5 s

4.025 s

5.4 s

6.75 s

Step-by-Step Solution

Identify Initial Conditions: When an object is dropped from a moving body (like a balloon), it acquires the initial velocity of that body at the instant of release.

Initial velocity ( $u$ ) = $+12 \text{ m/s}$ (directed upwards).
Displacement ( $s$ ) = $-81 \text{ m}$ (taking upward direction as positive, the final position (ground) is $81 \text{ m}$ below the point of release).
Acceleration ( $a$ ) = $-g = -10 \text{ m/s}^2$ (acting vertically downwards).
The mass of the body ( $2 \text{ kg}$ ) does not affect the kinematics of free fall (neglecting air resistance) .

Apply Equation of Motion: Use the displacement-time relation: $s = ut + \frac{1}{2}at^2$ . $-81 = 12t + \frac{1}{2}(-10)t^2$ $-81 = 12t - 5t^2$
Solve the Quadratic Equation: Rearrange the terms to form $5t^2 - 12t - 81 = 0$ . Using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $t = \frac{12 \pm \sqrt{(-12)^2 - 4(5)(-81)}}{2(5)}$ $t = \frac{12 \pm \sqrt{144 + 1620}}{10} = \frac{12 \pm \sqrt{1764}}{10}$ $t = \frac{12 \pm 42}{10}$ Possible values: $t_1 = 5.4 \text{ s}$ and $t_2 = -3 \text{ s}$ . Since time cannot be negative, $t = 5.4 \text{ s}$ .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started