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Q.

A rocket fired vertically ascends from rest with a constant acceleration g/3 for 1 min. Its fuel is then all used and it continues to rise as a free body. What is the maximum height reached above the ground ? $(g = 10 {ms}^{- 2})$

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a

6 km

b

2 km

c

4 km

d

8 km

answer is D.

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Detailed Solution

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0 to A (accelerated motion)
$\begin{array}{l} v_{0} = 0 + a_{1} t_{1} = \frac{g}{3} \times 60 = 200 m / s \\ h_{1} = 0 + \frac{1}{2} a_{1} t_{1}^{2} = \frac{1}{2} \times \frac{g}{3} \times (60)^{2} = 6000 m \end{array}$
A to B (motion under gravity):
$0 = v_{0}^{2} - 2 a_{2} h_{2}$
$0 = (200)^{2} - 2 \times 10 \times h_{2} = 2000 m$
Maximum height attained by the rocket $= h_{1} + h_{2} = 8000 m$

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A rocket fired vertically ascends from rest with a constant acceleration g/3 for 1 min. Its fuel is then all used and it continues to rise as a free body. What is the maximum height reached above the ground ? (g=10 ms−2)