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(C) During the first phase $(t_1 = 60\,s)$: Acceleration $a = 10\,m/s^2$. Height $h_1 = \frac{1}{2}at_1^2 = \frac{1}{2} 	imes 10 	imes (60)^2 = 1800

Vedclass · Accepted Answer

Both $(a)$ and $(b)$.

Vedclass · Answer

(C) During the first phase $(t_1 = 60\,s)$: Acceleration $a = 10\,m/s^2$. Height $h_1 = \frac{1}{2}at_1^2 = \frac{1}{2} 	imes 10 	imes (60)^2 = 18000\,m = 18\,km$. Velocity $v_1 = at_1 = 10 	imes 60 = 600\,m/s$. During the second phase (free fall): Maximum additional height $h_2 = \frac{v_1^2}{2g} = \frac{600^2}{2 	imes 10} = \frac{360000}{20} = 18000\,m = 18\,km$. Total maximum height $H_{\max} = h_1 + h_2 = 18\,km + 18\,km = 36\,km$. For total time of flight: Using $s = ut + \frac{1}{2}at^2$ for the whole journey,where $s = 0$ (displacement from ground): $0 = v_1 t_2 - \frac{1}{2}g t_2^2$ (where $t_2$ is time after fuel finishes). $-18000 = 600 t_2 - 5 t_2^2 \implies t_2^2 - 120 t_2 - 3600 = 0$. Solving for $t_2$: $t_2 = \frac{120 + \sqrt{14400 - 4(1)(-3600)}}{2} = \frac{120 + \sqrt{28800}}{2} = 60 + 60\sqrt{2}\,s$. Total time $T = t_1 + t_2 = 60 + 60 + 60\sqrt{2} = (120 + 60\sqrt{2})\,s$. Thus,both $(a)$ and $(b)$ are correct.

$A$ rocket is fired vertically up from the ground with a resultant acceleration of $10\,m/s^2$. The fuel is finished in $1\,min$ and it continues to move up. Take $g=10\,m/s^2$.

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