Question

A person wearing a parachute jumps off a plane from a height of 2 km from the ground and falls freely for 20 m before his parachute opens. After his parachute opens if he continues to move uniformly with the velocity attained due to his freefall, the total time taken by the person to reach the ground is (Acceleration due to gravity = 10 ms$^{-2}$)

• A. 99 s
• B. 100 s
• C. 101 s
• D. 102 s

Hint:

Break down complex motion problems into simpler parts. Identify the type of motion in each part (e.g., freefall, uniform velocity) and apply the appropriate kinematic equations for each segment. The total time or distance is the sum of the values from each segment.

Answer 1

The Correct Option is C

The motion consists of two parts.
Part 1: Freefall for a distance of $h_1 = 20$ m.
We need to find the time taken ($t_1$) and the final velocity ($v$) for this part. Using the kinematic equation $v^2 = u^2 + 2gh$, with initial velocity $u=0$.
$v^2 = 0^2 + 2(10)(20) = 400$.
$v = \sqrt{400} = 20$ m/s. This is the uniform velocity for the second part of the journey.
To find the time $t_1$, use $v = u + gt$.
$20 = 0 + 10t_1 \implies t_1 = 2$ s.
Part 2: Uniform motion for the remaining distance.
Total height $H = 2$ km = 2000 m.
Remaining height $h_2 = H - h_1 = 2000 - 20 = 1980$ m.
The person travels this distance with a uniform velocity $v = 20$ m/s.
The time taken for this part ($t_2$) is given by distance/speed.
$t_2 = \frac{h_2}{v} = \frac{1980}{20} = 99$ s.
Total time taken to reach the ground is the sum of the times for both parts.
Total time $T = t_1 + t_2 = 2 \text{ s} + 99 \text{ s} = 101$ s.