Answer
$$h =\left(-16 t^{2}+300\sqrt{2}t\right) \mathrm{ft}$$
Work Step by Step
Given $v= 600$ ft/sec , $ \theta =45 $ and
$$ h=-16t^2+vt\sin \theta $$
Then
\begin{aligned}
h &=\left(-16 t^{2}+600 \times \sin 45^{\circ} \times t\right) \\ &=\left(-16 t^{2}+600 \times \frac{1}{\sqrt{2}} \times t\right) \\
&=\left(-16 t^{2}+300\sqrt{2}t\right) \mathrm{ft}
\end{aligned}