Question

A rocket is fired vertically up from the ground with a resultant vertical acceleration of $10\, m / s ^{2}$. 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 $g =10\, m / s ^{2}$ )

• 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 $=60\, s$ ) in which resultant acceleration is vertically upwards is $10\, m / s ^{2}$ will be
$h_{1}=0 \times 60+(1 / 2) \times 10 \times 60^{2}$
$=18000\, m$ ...(A)
And velocity acquired by it will be
$v=0+10 \times 60=600\, m / s$ ...(B)
Now after 1 minute the rocket moves vertically up with initial velocity of $600\, m / s$ 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}-2 gh _{2} $
$\Rightarrow h _{2}=18000\, m$
[as $g =10\, m / s ^{2}$] ...(C)
So from eq. (A) and (C) the maximum height reached by the rocket from the ground.
$H = h _{1}+ h _{2} $
$=18+18=36\, km$
(B) As after burning of fuel the initial velocity from Eq. (B) is $600\, m / s$ 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 $=60\, s$
after finishing of fuel, the rocket goes up for $60\, \sec$ i.e., $1$ minute more.