Answer
As $v$ increases without bound, the equation of the asymptote is $\theta = 45^{\circ}$
Work Step by Step
Let $~~h=6$
$\theta = arcsin \left(\sqrt{\frac{v^2}{2v^2+64h}}\right)$
$\theta = arcsin \left(\sqrt{\frac{v^2}{2v^2+(64)(6)}}\right)$
$\theta = arcsin \left(\sqrt{\frac{v^2}{2v^2+384}}\right)$
As $v$ increases, the value of ${\frac{v^2}{2v^2+384}}$ approaches $\frac{1}{2}$. We can find the equation of the asymptote:
$\theta = arcsin \left(\sqrt{\frac{1}{2}}\right)$
$\theta = arcsin \left(0.707\right)$
$\theta = 45^{\circ}$
As $v$ increases without bound, the equation of the asymptote is $\theta = 45^{\circ}$
