A rocket takes off from a tower. For the first 40 seconds, the engines of the rocket provide enough power to launch it vertically. \( a(t)=3 t^{2}+6 \) \( \mathrm{v}(10)=2,00 \mathrm{ft} / \mathrm{sec} \) \( \mathrm{s}(10)=13,000 \mathrm{ft} \) What was the height of the tower? \( \mathrm{h}=[ \) ? ft

To find the height of the tower, we first need to determine the rocket's height at \( t = 0 \) seconds. We know the rocket's position function \( s(t) \) can be derived from the acceleration \( a(t) = 3t^2 + 6 \) by integrating. First, let’s find the velocity \( v(t) \) by integrating \( a(t) \): \[ v(t) = \int(3t^2 + 6) dt = t^3 + 6t + C_1 \] Using the given \( v(10) = 2.00 \, \text{ft/sec} \): \[ v(10) = 10^3 + 6(10) + C_1 = 1000 + 60 + C_1 = 1060 + C_1 \] Setting \( 1060 + C_1 = 2 \), we find \( C_1 = 2 - 1060 = -1058 \). The velocity function is: \[ v(t) = t^3 + 6t - 1058 \] Next, we find the position function \( s(t) \): \[ s(t) = \int(t^3 + 6t - 1058) dt = \frac{t^4}{4} + 3t^2 - 1058t + C_2 \] Using the given \( s(10) = 13000 \): \[ s(10) = \frac{10^4}{4} + 3(10^2) - 1058(10) + C_2 = 2500 + 300 - 10580 + C_2 \] \[ s(10) = 2500 + 300 - 10580 + C_2 = 13000 \] This simplifies to \( -8300 + C_2 = 13000 \), so \( C_2 = 21300 \). Thus, the position function is: \[ s(t) = \frac{t^4}{4} + 3t^2 - 1058t + 21300 \] Finally, we find the height of the rocket at \( t = 0 \): \[ s(0) = \frac{0^4}{4} + 3(0^2) - 1058(0) + 21300 = 21300 \, \text{ft} \] Therefore, the height of the tower \( h \) is \( 21300 \, \text{ft} \).