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

(i) The distance travelled by the rocket during burning of fuel (1 minute = 60 s) in which resultant acceleration is vewrtically upwards and is `10 ms^(-2)` will be
`h_(1)=0xx60+(1//2)xx10xx60^(2)=18000 m =18 kn` and velocity acquired by it will be
`v=0+10xx60=600 ms^(-1)`.
Now after 1 min the rocket moves vertically up with initial velocity of `600 ms^(-1)` and acceleration due to gravity opposes its motion. So, it will go to a height `h_(2)` from this point, till is velocity becomes zero such that `0=(600)^(2)-2gh_(2)` or `h_(2)=18000 m = 18 km (g=10 ms^(-2))`
So the maximum height reached by the rocket from the ground, `H = h_(1)+h_(2)=18+18=36 km`
(ii) As after burning of fuel the initial velocity attained will be `600 ms^(-1)` and gravity opposes the motion of rocket, so from 1st equation of motion time taken by it till it velocity v = 0
`0=600-"gt " rArr t = 60 s`