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
Free fall for $3$ s: $v = 30$ m/s, and $h_1 = \dfrac{1}{2}(10)(3)^2 = 45$ m.
With the parachute, from $30$ m/s to $4$ m/s at $2\ \text{m/s}^2$ retardation:
$$4^2 = 30^2 - 2(2)h_2 \;\Rightarrow\; h_2 = \frac{900 - 16}{4} = 221\ \text{m}$$
Total height $= 45 + 221 = 266$ m.
Solution by Sreeraj P, M.Sc Physics