A rocket is launched at an angle `53^(@)` to the horizontal with an initial speed of `100ms^(-1)`. It moves along its initial line of motion with an acceleration of `30ms^(-2)` for `3` seconds. At this time its engine falls & the rocket proceeds like a free body. Find:
(i) the maximum altitude reached by the rocket
(ii) total time of flight.
(iii) the horizontal range. `[sin 53^(@) = 4//5]`

Answer

(i) `1503.2 m` (ii) `35.54sec` (iii) `3970.56 m`

Explanation

corrdinate of the orcket when engine falls
`r = ut +(1)/(2)at^(2)`
`r = 100 xx 3 +(1)/(2) xx 30 xx (3)^(2)`

`r = 300 +135 rArr r = 435 m`
`x = r cos theta
x = 435xx 0.6 rArr x = 261 m`
`y = r sin theta rArr y = 435 xx 0.8 rArr y = 348m`
velocity of rocket at this instant
`v = u +at rArr v = 100 +30 xx 9 rArr v = 190 m//s`
(i) Maximum height attained
`= 348 +(u_(y)^(2))/(2g) 348 +((190 xx 0.8)^(2))/(2xx10)`
maximum height attained
`= 348 +1155.2 = 348 +1155.2`
maximum height attained `= 1503.2 m`
Time fo flight
`y = y_(0) +u_(y)t +(1)/(2) a_(y)t^(2)`
`0 = 348 +190 xx 0.8 xx t +(1)/(2) xx (-10)t^(2)`
`5t^(2) -152 t - 348 = 0 rArr t = 32.54s`
Time of flight `= 32.54 +3 = 35.54 s`
(iii) Horizontal range `= 261 +R`
horizontal range `= 261 +(190 xx 0.6) xx 32.54`
horizontal range `= 261 +(u cos theta) t = 3970.56m`