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 min and it continues to move up. (a) What is the maximum height reached? (b) Afte2r how much time from then will the maximum height be reached?(Take `g= 10 m//s^2`)

Answer 1

Correct Answer - A::B::C
(a) The distance travelled by the rocket in 1 min
`(=60s)` in which resultant acceleration is vetically
upwards and `10m//s^2` will be
`h_1=(1//2)xx10xx60^2=18000m=18km…(i)`
and velocity acquired by it will be
`v=10xx60=600m//s...(ii)`
Now, after 1min the rocket moves vetically up
with velocity of `600m//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)^2-2gh_2`
or `h_2=18000m["as" g=10m//s^2]....(iii)`
`=18km`
So, from Eqs. (i) and (iii) 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. (ii) is `600m//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=60s`
i.e. after finishing fuel the further goes up
for 60 s, or 1min.