The distance travelled by the rocket in $1 \mathrm{min}$ $(=60 s)$ in which resultant acceleration is vetically upwards and $10 \mathrm{m} / \mathrm{s}^{2}$ will be $h_{1}=(1 / 2) \times 10 \times 60^{2}=18000 \mathrm{m}=18 \mathrm{km} \ldots(i)$

and velocity acquired by it will be $v=10 \times 60=600 m / s \ldots(i i)$

Now, after $1 \mathrm{min}$ the rocket moves vetically up with velocity of $600 \mathrm{m} / \mathrm{s}$ and acceleration due to gravity opposes its motion. So, it will go to a height $h_{2}$ till its velocity becomes zero such that $0=(600)^{2}-2 g h_{2}$

or $h_{2}=18000 \mathrm{m}\left[\mathrm{as} g=10 \mathrm{m} / \mathrm{s}^{2}\right] \ldots .(\text { iii })$

$=18 k m$

So, from Eqs. (i) and (iii) the maximum height reached by the rocket from the ground $h=h_{1}+h_{2}=18+18=36 k m$

As after burning of fuel the initial velocity from Eq. ( i. ) is $600 \mathrm{m} / \mathrm{s}$ and gravity opposes the motion of rocket, so the time taken by it to reach the maximum height (for which $v=0$ ). $0=600-g t$

or $t=60 s$

i.e. after finishing fuel the further goes up for $60\; s$, or $1 \mathrm{min}$.