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A rocket is fired vertically up from the ground with a resultant acceleration of $10\,m / s ^2$. The fuel is finished in $1 min$ and it continues to move up $\left(g=10\,m / s ^2\right)$
the time from initial in which rocket is again at ground is $(120+60 \sqrt{2})\,s$
the maximum height reached by rocket from ground is $36\,km$
Only $(a)$ and $(b)$
the maximum height reached by rocket from ground is $18\,km$
Solution
(c)
At $60\,s$
$h_1=\frac{1}{2} \times 10 \times(60)^2=18000\,m$
$v_1=10 \times 60=600\,m / s$
After that
$\therefore \quad H_{\max }=h_1+h_2=36000 m =36\,km$
For time
$-18000=600 t-5 t^2$
or $t^2-120 t-3600=0$ or
$t =\frac{120+\sqrt{14400+14400}}{2}$
$=\frac{120+120 \sqrt{2}}{2}=60+60 \sqrt{2}$
$\therefore \quad \text { Total time }=60+(60+60 \sqrt{2})=(120+60 \sqrt{2})\,s$
