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(C) Step $1$: Calculate the velocity and height at the end of the powered flight $(t = 5 \,s)$.The initial velocity $u = 0$, acceleration $a = 20 \,m/

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$50 \sqrt{6} \,m/s$

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(C) Step $1$: Calculate the velocity and height at the end of the powered flight $(t = 5 \,s)$.The initial velocity $u = 0$, acceleration $a = 20 \,m/s^2$, and time $t = 5 \,s$.The velocity at $t = 5 \,s$ is $V_{max} = u + at = 0 + 20 	imes 5 = 100 \,m/s$.The height reached at $t = 5 \,s$ is $S = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} 	imes 20 	imes 5^2 = 250 \,m$.Step $2$: Calculate the final speed when it hits the ground.After $t = 5 \,s$, the rocket is in free fall under gravity $(g = 10 \,m/s^2)$.Using the work-energy theorem or kinematic equations, the total energy at the peak height is conserved, or we can use $V^2 = u^2 + 2aS$.Here, the rocket starts from height $S = 250 \,m$ with an initial upward velocity $V_{max} = 100 \,m/s$.When it hits the ground, the final velocity $V$ is given by $V^2 = V_{max}^2 + 2gS$.$V^2 = (100)^2 + 2 	imes 10 	imes 250$.$V^2 = 10000 + 5000 = 15000$.$V = \sqrt{15000} = \sqrt{2500 	imes 6} = 50 \sqrt{6} \,m/s$.

$A$ rocket moves straight upward with zero initial velocity and with an acceleration of $20 \,m/s^2$. It runs out of fuel and stops accelerating at the end of $5^{th} \,sec$. It reaches a maximum height and falls back to the earth. The speed when it hits the ground is (Take $g = 10 \,m/s^2$):

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