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Question

(A) The distance travelled by the rocket in $1\ min$ $(= 60\ s)$ during which the resultant acceleration is vertically upwards and $10\ m/s^2$ is give

Vedclass · Accepted Answer

$36\ km, 1\ min$

Vedclass · Answer

(A) The distance travelled by the rocket in $1\ min$ $(= 60\ s)$ during which the resultant acceleration is vertically upwards and $10\ m/s^2$ is given by $h_1 = \frac{1}{2} 	imes 10 	imes 60^2 = 18000\ m = 18\ km$. The velocity acquired by the rocket at the end of $1\ min$ is $v = u + at = 0 + 10 	imes 60 = 600\ m/s$. After the fuel is finished, the rocket moves vertically up under gravity. The acceleration is $-g = -10\ m/s^2$. It reaches a maximum height $h_2$ when its final velocity becomes $0$. Using $v^2 = u^2 + 2as$, we get $0 = (600)^2 - 2 	imes 10 	imes h_2$, which gives $h_2 = \frac{360000}{20} = 18000\ m = 18\ km$. The total maximum height reached from the ground is $h = h_1 + h_2 = 18\ km + 18\ km = 36\ km$. To find the time taken after the fuel is finished to reach the maximum height, we use $v = u + at$, where $v = 0$, $u = 600\ m/s$, and $a = -10\ m/s^2$. Thus, $0 = 600 - 10t$, which gives $t = 60\ s = 1\ min$.

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