Q 11-02-036NEETJEE MainHard
A skydiver falls freely from rest for $3$ s. Then her parachute opens and she decelerates uniformly at $2\ \text{m/s}^2$, reaching the ground with a speed of $4$ m/s. From what height did she start? ($g = 10\ \text{m/s}^2$)
Answer: (A) $266$ m
Concept. Split into free fall and uniform deceleration and add the heights.
Step 1. Free fall ($3$ s). $v = 30$ m/s, $h_1 = \dfrac12(10)(9) = 45$ m.
Step 2. Parachute phase.
$$4^2 = 30^2 - 2(2)h_2 \;\Rightarrow\; h_2 = \frac{900 - 16}{4} = 221\ \text{m}$$
Step 3. Total. $45 + 221 = 266$ m.
Answer. $266$ m (Option A).
Solution by Sreeraj P, M.Sc Physics