A rocket is fired vertically up from the ground with a resultant vertically acceleration of `10m//s^(2)`. The fuel is finished in 1 minute and it continues to move up.
(a) What is the maximum height reached?
(b) After finishing fuel, calculate the time for which it continues its upwards motion. (Take g= `10 m//s^(2)`).

Explanation

(a) The distance travelled by the rocket during burning interval (1 minute = 60 s) in which resultant acceleration is vertically upwards and `10 m//s^(2) ` will be `h_1 = 0 xx 60 + (1//2) xx 10 xx 60^(2) = 18000 `m = 18 km and velocity acquired by it will be `v= 0 + 10 xx 60 = 600 m//s`
Now after 1 minute the rocket moves vertically up with initial velocity of 600 m/s and acceleration due to gravity opposes its motion. So, it will go to a height `h_2` from this point, till its velocity becomes zero such that `0 = (600)^(2) - 2gh_2 or h_2 = 18000` m = 18 km [ g= 10 `m//s^(2)`]
So 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 600 m/s and gravity opposes the motion of rocket , so from `1^(st)`
`0 = 600 - "gt" rArr t = 60 s`