How to solve $-\frac{1}{2}gt^2\sin \theta=x$ when $x$ equals $0$

Question

This given equation, $-\dfrac{1}{2}gt^2\sin \theta=x$ should describe the path of a ball rolling up an incline with initial speed $v_0$.

The question is: how long will the ball take to return to its starting point?

It is obvious that the equation must be solved for $x=0$, which has two solutions, according to my solution manual: for $t=0$ and for $t=\dfrac{2v_0}{g\sin \theta}$

However, I do not understand how the second solution is obtained?

Can someone help me? Thanks!

Answer 1

The motion equation should be

$$x=-\frac 12gt^2\sin (\theta)+v_0t $$

and for $x=0,$ we get the desired solutions.

Answer 2

The correct expression should be

$$v_0t-\frac{1}{2}gt^2\sin \theta=x$$

and thus for $x=0$

$$v_0t-\frac{1}{2}gt^2\sin \theta=0\implies t=0 \quad \lor \quad t=\dfrac{2v_0}{g\sin \theta} $$