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(C) Step $1$: Calculate the height reached during the powered flight $(S_1)$.Given acceleration $a = 20\, m/s^2$,time $t = 5\, s$,and initial velocity

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$750$

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(C) Step $1$: Calculate the height reached during the powered flight $(S_1)$.Given acceleration $a = 20\, m/s^2$,time $t = 5\, s$,and initial velocity $u = 0\, m/s$.$S_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} 	imes 20 	imes 5^2 = 250\, m$.Step $2$: Calculate the velocity at the end of the powered flight $(v)$.$v = u + at = 0 + 20 	imes 5 = 100\, m/s$.Step $3$: Calculate the additional height reached after the engine is switched off $(S_2)$.At this point,the rocket is under gravity ($g = 10\, m/s^2$ downward). The final velocity at maximum height is $0\, m/s$.Using $v_f^2 = v^2 - 2gS_2$,we get $0 = 100^2 - 2 	imes 10 	imes S_2$.$S_2 = \frac{10000}{20} = 500\, m$.Step $4$: Total height $H = S_1 + S_2 = 250 + 500 = 750\, m$.

$A$ rocket is fired upward from the earth's surface such that it creates an acceleration of $20\, m/s^2$. If after $5\, s$ its engine is switched off,the maximum height of the rocket from the earth's surface would be......$m$

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