A rocket is fired vertically up from the ground with a resultant vertical acceleration of $10 \mathrm{m}/{\mathrm{s}}^2$. The fuel is finished in $1 \min$ and it continues to move up. What is the maximum height reached?

detailed solution

Correct option is C

The distance travelled by the rocket in  1 min (=60 s) in which resultant acceleration is vertically upwards and 10 m/s2 will be h1=(1/2)×10×602=18000 m=18 km----(i)and velocity acquired by it will bev=10×60=600 m/s  ----(ii)Now, after 1 min the rocket moves vertically up with velocity of 600 m/s and acceleration due to gravity opposes its motion. So, it will go to a height h2 till its velocity becomes zero such that0=(600)2-2gh2 h2=18000 m as g=10 m/s2   ----(iii) =18 kmSo, from Eqs. (i) and (iii) the maximum height reached by the rocket from the groundh=h1+h2=18+18=36 km