since it told me to express the maximum horizontal distance, I thought it would make sense to make y = 0, because the ball reaches its maximum horizontal distance when the ball finally drops back down to ground level. So I continued to solve for x algebraically, but I ended up with a different answer. So can someone tell me what I did wrong?
The correct answer is:
my working out:
Expanding $\tan x$ you can see that
$$x = -\frac{v^2\frac{\sin \theta}{\cos \theta}\cos^2\theta}{5} = -\frac{v^2\sin \theta\cos \theta}{5}$$
$$\implies x = -\frac{v^2\sin 2\theta}{10}$$