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`)

Options

A. 36 km, 1 min
B. 6 km, 1 min
C. 36 km, 1 sec
D. 36 km, 1 sec

Explanation

(A) The distance travelled by the rocket during burning interval (1 minute = 60 s) in which resultant acceleration is vertically upwards is `10 m//s^(2)` will be
` h_(1) = 0 xx 60 + (1//2) xx 10 xx 60^(2) = 18000 ` m ….(A)
And velocity acquired by it will be
`v = 0 + 10 × 60 = 600 m/s` .....(B)
Now after 1 minute the rocket moves vertically up
with initial velocity of 600 m/s and acceleration due
to gravity oppose its motion. So, it will go to a height
`h^(2)` till its velocity becomes zero that
` 0 = (600)^(2) - 2gh_(2) rArr h_(2) = 18000 ` m
[as `g = 10 m//s^(2)]` ....(C)
So from eq. (A) and (C) 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 Eq. (B) is 600 m/s and gravity opposes the motion of rocket, so from 1st equation of motion time taken by
it to reach the maximum height (for which v = 0)
0 = 600 – gt, i.e.t = 60 s
after finishing of fuel, the rocket goes up for 60 sec
i.e., 1 minute more.
Hence correct answer is (A)