A rocket takes off from a tower. For the first 40 seconds, the engines of the rocket provide enough power to launch it vertically. $$ a(t)=3 \mathrm{t}^{2}+6 $$ $$ \mathrm{v}(10)=2,000 \mathrm{ft} / \mathrm{sec} $$ $$ \mathrm{s}(10)=13,000 \mathrm{ft} $$ What was the height of the tower? h $$ =[?] \mathrm{ft} $$

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We are given the acceleration function
$$
a(t)=3t^2+6.
$$
First, we integrate the acceleration to find the velocity. Since
$$
v(t)=\int a(t)\,dt,
$$
we have
$$
v(t)=\int \left(3t^2+6\right)dt=t^3+6t+C_1.
$$
We know that
$$
v(10)=2000\text{ ft/sec},
$$
so by substituting $$ t=10 $$ we get
$$
10^3+6\cdot10+C_1=2000.
$$
That is,
$$
1000+60+C_1=2000 \quad\Longrightarrow\quad C_1=2000-1060=940.
$$
Thus, the velocity function becomes
$$
v(t)=t^3+6t+940.
$$

Next, we integrate the velocity to find the position function:
$$
s(t)=\int v(t)\,dt=\int \left(t^3+6t+940\right)dt=\frac{1}{4}t^4+3t^2+940t+C_2.
$$
The rocket takes off from the top of the tower, so the initial height $$ s(0) $$ is the tower's height $$ h $$:
$$
s(0)=C_2=h.
$$

We are given that at $$ t=10 $$ seconds, the position is
$$
s(10)=13000\text{ ft}.
$$
Substitute $$ t=10 $$ into $$ s(t) $$:
$$
s(10)=\frac{1}{4}(10)^4+3(10)^2+940\cdot10+h.
$$
Calculating each term:
$$
\frac{1}{4}(10)^4=\frac{1}{4}\cdot10000=2500,
$$
$$
3(10)^2=3\cdot100=300,
$$
$$
940\cdot10=9400.
$$
Therefore,
$$
s(10)=2500+300+9400+h=12200+h.
$$
Setting this equal to 13000 ft gives
$$
12200+h=13000.
$$
Solving for $$ h $$ yields
$$
h=13000-12200=800\text{ ft}.
$$

Thus, the height of the tower is $$ \boxed{800\text{ ft}} $$.
The height of the tower is 800 feet.

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