Question

A rocket is fired vertically up from the ground with an acceleration 10 `m//s^(2)`. If its fuel is finished after 1 minute then calculate -
(a) Maximum velocity attained by rocket in ascending motion.
(b) Height attained by rocket fuel is finished.
(c ) Time taken by the rocket in the whole motion.
(d) Maximum height attained by rocket.

Answer 1

Point O `to` Point of projection
Point A `to` Point up to which fuel is consumed.
Point B `to` Highest point of path
`h_1 = 0xx 60 + (1)/(2) xx 10 xx (60)^(2)`
` = 18` km
`v_A =0 + 10 xx 60 = 600 m//s `
`O = v_A^(2) - 2gh_2 rArr h_2 = (v_(A)^(2))/(2g) = ((600)^(2))/(20)= 18 ` km.
maximum height from ground = `18 + 18 = 36` km.
time taken from A to B : `to` O = 600 - gt `rArr t= 60 sec`.
time taken in coming down to earth -
`36000 = (1)/(2) "gt"^(2) rArr t = 60 sqrt2 sec`.
`therefore ` Total time = `60 + 60+ 60 sqrt2 = 60 (2+ sqrt2)s`.
`" " = ( 2 + sqrt2) min`.