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`)

Explanation

(a) The distance travelled by the rocket during burning interval (1minute=6s) in which resultant acceleration is vertically upwards and `10 m//s^(2)` will be `h_(1)=0xx60+(1//2)xx10xx60^(2)=18000 m=18 km` and velocity acquired by it will be `v=0+10xx60=600m//s`
Now after 1 minute the rocket moves vertically up with initial velocity of 600 m/s and accelration 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=10ms^(-2)|`
(b) As after burning of fuel the initial velocity 600 m/s and gravity opposes the motion of rocket, so from `1^(st)` equation of motion time taken by it till it velocity v=0
`=0600-gtrArr t=60s`