A rocket is fired vertically upward from the ground with a resultant vertical acceration of `5 m//s^(2)`. The fuel is finished in `100 s` and it continues to move up. After how much time from then will the maximum height be reached and what is the maximum height?

Explanation

`t=0` to `t=100s`, motion under constant acceleration,` tgt100 s`, motion under gravity

`O` to `A`(accelerated motion)
`v_(0)=0+a_(1)t_(1)=5xx100=500 m//s`
`h_(1)=01/2a_(1)t_(1)^(2)=1/2xx5xx(100)^(2)=2.5xx10^(4) m`
`A` to `B` (motion under gravity)
`0=v_(0)-a_(2)t_(2)=500-10t_(2)impliest_(2)=50 s`
`0=v_(0)^(2)-2a_(2)h_(2)=(500)^(2)-2xx10xxh_(2)`
`h_(2)=1.25xx10^(4) m` ltbegt Maximum height attained`=h_(1)+h_(2)=3.75xx10^(4) m`
Time from `A` to `B =50 S`