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(D) Given acceleration $a = 19.6 \,m/s^2 = 2g$ (where $g = 9.8 \,m/s^2$).Phase $1$: Motion with engine on for $t_1 = 5 \,s$.The velocity attained at $

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$735$

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(D) Given acceleration $a = 19.6 \,m/s^2 = 2g$ (where $g = 9.8 \,m/s^2$).Phase $1$: Motion with engine on for $t_1 = 5 \,s$.The velocity attained at $t_1 = 5 \,s$ is $v = u + at_1 = 0 + (19.6)(5) = 98 \,m/s$.The height reached during this phase is $h_1 = ut_1 + \frac{1}{2}at_1^2 = 0 + \frac{1}{2}(19.6)(5)^2 = 245 \,m$.Phase $2$: Motion under gravity after engine is switched off.The rocket moves upward with initial velocity $u_2 = 98 \,m/s$ and acceleration $a_2 = -g = -9.8 \,m/s^2$.At maximum height,final velocity $v_2 = 0$.Using $v_2^2 = u_2^2 + 2a_2h_2$,we get $0 = (98)^2 + 2(-9.8)h_2$.$19.6h_2 = 9604 \Rightarrow h_2 = 490 \,m$.Total height $H = h_1 + h_2 = 245 + 490 = 735 \,m$.

$A$ rocket is fired upward from the earth's surface such that it creates an acceleration of $19.6 \,m/s^2$. 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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