Pretty sure I have every part of this problem solved but my final answer concerns me as it seems a little high.

$\displaystyle a(t)=(3-t)42-9.8m/s^2$ for 0<=t<=3

and $\displaystyle a(t)=-9.8m/s^2$ for t>3

I should find the maximum altitude and time in which it gets there. My answer is 56s and 18565.18meters.

**additional info**

initial velocity ($\displaystyle v_i$ is 366.4m/s)

initial position ($\displaystyle x_i$) is 0

$\displaystyle \int (3-t)42-9.8 dx=v(t_i)=126t-42t-9.8t+v_i$

after taking integral of v(t) $\displaystyle x(t_i)=-7t^3+63t^2+366.4t+x_i$

**

$\displaystyle x(t_i)$= position fxn until 3 seconds.

$\displaystyle x(t_i)=-7t^3+63t^2+366.4t=1477.2m$ after 3 seconds of burn

$\displaystyle x(t_m)$=position fxn after 3s initial burn.

$\displaystyle x(t_m)=\frac{9.88t^2}{2}+555.4t+1477.2$

$\displaystyle x(t_m)=\frac{9.88t^2}{2}+555.4t+1477.2=0$ at t=56.2145s

$\displaystyle x'(t_m)$where t=56.2145s

$\displaystyle x'(56.2145)=17087.98m$

17087.98+1477.2=18565.18meters

Is it me or is that pretty dang high?