Question

A rocket is fired vertically up from the ground with a resultant vertical acceleration of $10m/s^2$. The fuel is finished in $1$ minute and it continues to move up. $(1)$ the maximum height reached. $(2)$ After how much time from then will the maximum height be reached (Take $g=10m/s^2$ )

• A. 36 km , 1 min
• B. 6 km , 1 min
• C. 36 km , 1 sec
• D. 36 km , 1 sec

Answer 1

Answer: (A)

The distance travelled by the rocket during burning interval ( $1$ minute $=60s$ ) in which resultant acceleration is vertically upwards is $10m/s^2$ will be
$h_1=0\times 60+(1/2)\times 10\times 60^2$
$=18000m$ ...(A)
And velocity acquired by it will be
$v=0+10\times 60=600m/s$ ...(B)
Now after 1 minute the rocket moves vertically up with initial velocity of $600m/s$ and acceleration due to gravity oppose its motion.
So, it will go to a height $h_2$ till its velocity becomes zero that
$0=(600{)}^2-2g{h}_2$
$\Rightarrow h_2=18000m$
[as $g=10m/s^2$] ...(C)
So from eq. (A) and (C) the maximum height reached by the rocket from the ground.
$H=h_1+h_2$
$=18+18=36km$
(B) As after burning of fuel the initial velocity from Eq. (B) is $600m/s$ and gravity opposes the motion of rocket, so from 1st equation of motion time taken by it to reach the maximum height (for which $v=0$ )
$0=600-gt$ i.e.t $=60s$
after finishing of fuel, the rocket goes up for $60\sec$ i.e., $1$ minute more.