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
Concept. Split into the powered phase (acceleration $5\ \text{m/s}^2$) and the free rise under gravity.
Step 1. Powered phase.
$$v = 5 \times 20 = 100\ \text{m/s},\qquad h_1 = \frac12(5)(20)^2 = 1000\ \text{m}$$
Step 2. Free rise. $h_2 = \dfrac{100^2}{2 \times 10} = 500$ m.
Step 3. Maximum height. $1000 + 500 = 1500$ m.
Answer. $1500$ m (Option D).
Solution by Sreeraj P, M.Sc Physics