Q 11-02-035NEETJEE MainMedium
A rocket is launched vertically upward from rest with a constant acceleration of $5\ \text{m/s}^2$. After $20$ s its fuel runs out and it continues as a free body. The maximum height it reaches is ($g = 10\ \text{m/s}^2$)
Answer: (D) $1500$ m
Powered phase ($20$ s):
$$v = 5 \times 20 = 100\ \text{m/s}, \qquad h_1 = \frac{1}{2}(5)(20)^2 = 1000\ \text{m}$$
Free phase (rising under gravity until $v = 0$):
$$h_2 = \frac{v^2}{2g} = \frac{10000}{20} = 500\ \text{m}$$
Maximum height $= 1000 + 500 = 1500$ m.
Solution by Sreeraj P, M.Sc Physics