Paratrooper jumps... opens parachute after 2 s... initial height… — Solution | JEE Physics

MCQMediumJEE 2026Motion in a Straight Line

Paratrooper jumps... opens parachute after $2 \, \text{s}$ ... initial height is _____ $\text{m}$ .

A
$82.5$
B
$92.5$
C
$62.5$
D
$20$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The motion is divided into segments. The paratrooper opens the parachute after $2 \, \text{s}$ . During the first segment, acceleration is due to gravity. After that, use the given retarded motion from the solution.

Find: The initial height.

In the first $2 \, \text{s}$ ,

h_1 = \frac{1}{2}(10)(4) = 20

so the speed becomes $v = 20 \, \text{m/s}$ .

For the next segment,

v^2 - u^2 = 2as

25 - 400 = 2(-3)h_2

h_2 = 62.5

Completion of Total Height

From the extracted solution,

h_3 = 10

Therefore, total height is

20 + 62.5 + 10 = 92.5

So the initial height is $92.5 \, \text{m}$ and the correct option is B.

Common mistakes

Treating the entire motion as a single constant-acceleration segment. This is wrong because the solution clearly splits the motion into different phases. Instead, calculate each segment separately and then add the distances.
Using the wrong initial velocity in the second segment. After the first $2 \, \text{s}$ , the speed is $20 \, \text{m/s}$ , not zero. Use this as the starting speed for the next kinematic equation.
Ignoring the final segment $h_3$ . This gives only a partial height. After finding $h_1$ and $h_2$ , also include $h_3 = 10$ before computing the total.

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Answer

Step-by-step solution

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