A cannon ball is shot from the ground with velocity $v$ at an angle $\theta$. Its position as it moves in the horizontal direction is $x(t)=vt \cos\theta$, while its vertical position is given by $y(t)=vt \sin\theta−4.9t^2$. Find the angle $\theta$ that will maximize the horizontal distance traveled by the cannon ball before striking the ground.
I am not entirely sure as to how to approach this problem.. so please correct me if I am wrong. I assume the angle which maximizes the horizontal distance is $45$ degrees, I just don't know how to fully prove it.
Here is what I have so far:
max height at time $t$:
$$ y(t)=v \sin\theta-4.9t^2 $$
$$ y'(t)=v \sin\theta-9.8t=0 $$
$$ v\sin\theta=9.8t $$
$$ \frac{v\sin\theta}{9.8}=t $$
So (I think) at time $t=\frac{\sin\theta}{9.8}$ the ball with be at its maximum height.
now I am unsure as to how to get its maximum horizontal distance from this information, assuming what I have so far is correct.
Note that your first line of analysis has omitted a factor of $t$ in the expression for $y(t)$. Your expression for $y'(t)$ is correct, however.
One way to proceed is to observe that (on level ground, which we'll somewhat recklessly assume is the case here) the ball hits the ground at twice the time it takes to reach its peak, so that the total travel time is $2 \times \frac{v \sin\theta}{9.8} = \frac{v \sin\theta}{4.9}$. If we plug that into our expression for $x(t)$, we get
\begin{align} x\left(\frac{v \sin\theta}{4.9}\right) & = v\left(\frac{v \sin\theta}{4.9}\right)\cos\theta \\ & = \frac{v^2 \sin\theta \cos\theta}{4.9} \\ & = \frac{v^2 \sin 2\theta}{9.8} \end{align}
where the last line is due to the identity $\sin 2\theta = 2\sin\theta\cos\theta$. If we take $v$ as given, then the only parameter of interest is $\theta$, and we can see that $\sin 2\theta$ reaches a maximum of $1$ whenever $2 \theta = 90$ degrees, meaning that $\theta$ would have to be $45$ degrees.