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(A) Given, resultant vertical acceleration $a = 10 \,m/s^2$, time $t = 1 \,min = 60 \,s$, and initial velocity $u = 0$.Height reached during the power

Vedclass · Accepted Answer

$36.4$

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(A) Given, resultant vertical acceleration $a = 10 \,m/s^2$, time $t = 1 \,min = 60 \,s$, and initial velocity $u = 0$.Height reached during the powered phase $(h_1)$:$h_1 = ut + \frac{1}{2}at^2 = 0 	imes 60 + \frac{1}{2} 	imes 10 	imes (60)^2 = 18000 \,m = 18 \,km$.Velocity at the end of the powered phase $(v)$:$v = u + at = 0 + 10 	imes 60 = 600 \,m/s$.After the fuel is exhausted, the rocket moves under gravity $(a = -g = -9.8 \,m/s^2)$ until its velocity becomes zero.Height reached during the unpowered phase $(h_2)$:Using $v_f^2 - v^2 = 2ah_2$, where $v_f = 0$:$0 - (600)^2 = 2(-9.8)h_2 \Rightarrow h_2 = \frac{360000}{19.6} \approx 18367.3 \,m \approx 18.4 \,km$.Total maximum height $H = h_1 + h_2 = 18 \,km + 18.4 \,km = 36.4 \,km$.

$A$ rocket is fired vertically from the ground with a resultant vertical acceleration of $10 \,m/s^2$. The fuel is exhausted in $1 \,min$ and it continues to move up. What is the maximum height reached (in $\,km$)?

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