(a) The distance travelled by the rocket during burning interval (1 minute = 60 s) in which resultant acc. Is vertically upwards and `10 m//s^(2)` will be
`h_(1) = 0 xx 60 + (1//2) xx 10 xx 60^(2) = 18000 m`....(i)
and velocity acquired by it will be
`v = 0 + 10 xx 60 = 600 m//s`....(ii)
Now after 1 minute the rocket moves vertically up with initial velocity of `600 m//s` and acc. due to gravity oppose its motion, So, it will go to a height `h_(2)` till its velocity becomes zero such that
`0 = (600)^(2) - 2 gh_(2)`
or `h_(2) = 18000 m` [as `g = 10 m//s^(2)`] ....(iii)
so from Eqns. (i) and (iii) 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 Eqn. (ii) is `600 m//s` and gravity opposes the motion of rocket, so from 1 st equation of motion time taken by it to reach the maximum height (for which `v = 0`)
`0 = 600 - g t`
i.e., `t = 60 s`
i.e., after finishing fuel the rocket goes up for `60 s`, i.e., 1 minute more.
