## Firework Rocket

#### Problem

A firework rocket is fired vertically upwards with a constant acceleration of 4 ms^{-2} until the chemical fuel expires. Its ascent is then slowed by gravity until it reaches a maximum height of 138 metres.

Assuming no air resistance and taking `g`=9.8 ms^{-2}, how long does it take to reach its maximum height?

#### Solution

During acceleration phase (as fuel burns), `a`=4, `u`=0, let `v=w`.

`v`=`u`+`at`: `w`=4`t`_{1} (1)`v`^{2}=`u`^{2}+2`as`: w^{2}=8`s`_{1} (2)

During deceleration phase (fuel expired), `a`=-9.8, `u=w`, `v`=0.

`v`=`u`+`at`: 0=`w`9.8`t`_{2}, `w`=9.8`t`_{2} (3)`v`^{2}=`u`^{2}+2`as`: 0=`w`^{2}19.6`s`_{2}, `w`^{2}=19.6`s`_{2} (4)

Equating (1) and (3), 4`t`_{1}=9.8`t`_{2}, `t`_{2}=(20/49)`t`_{1}, so total time to reach maximum height, `t`_{1}+`t`_{2}=(69/49)`t`_{1}.

Equating (2) and (4), 8`s`_{1}=19.6`s`_{2}, `s`_{2}=(20/49)`s`_{1}, and as`s`_{1}+`s`_{2}=(69/49)`s`_{1}=138, we get `s`_{1}=98.

Using `s`=`ut`+½`at`^{2} during acceleration phase, `s`_{1}=2`t`_{1}^{2}, `t`_{1}=(s_{1}/2)=7.

Hence time to reach maximum height is (69/49)7=69/7 seconds.