# What does the $y$-axis in quadratic equation graph when solving for the vertical displacement mean?

A projectile is shot from a cannon with an initial velocity of 20 $$\frac ms$$ from the base of an incline surface of angle 25°. The angle of projection with the incline surface is 10°. How far will the projectile go along the incline surface?

This isn't my question though but an example. I started off by trying to solve for the vertical displacement, $$\vec{S_y}$$, which I ended up with $$|\vec{S_y}| = \frac{(-4.9)\cdot(|\vec{S_y}|)^2}{(\tan^2 25°)\cdot(\cos^2 35°)\cdot20^2}+ \frac{(\sin 35°)\cdot |\vec{S_y}|}{(\tan 25°)\cdot(\cos 35°)}$$ I did the math and got $$|\vec{S_y}|$$ = 5.974$$\frac ms$$ or $$|\vec{S_y}|$$ = 0$$\frac ms$$. So far, so good until I decided to draw the graph. I got the equation above from the kinematic formula $$S = V_i \cdot \Delta t + \frac 12 \cdot a \cdot \Delta t^2$$ In a 'normal case', I would have 2 variables. The independant variable (in the x-axis) will be $$\Delta t$$ and the dependent variable (in the y-axis) will be $$S_y$$. But since it's being shot onto an incline surface and I need to relate both $$S_y$$ and $$S_x$$ (and I tried solving for $$S_y$$), I ended up with the equation having only 1 variable, $$S_y$$ (as seen in the equation above).

I made the graph anyway (the last picture) and got a parabola. I see that the x-intercept is what $$S_y$$ equals to. That means (or at least what I thought it means), the x-axis represents the vertical displacement, which is quite new to me.

What does the y-axis represent in this graph then? I do believe that it represents something. Thanks. In your last plot, $$x$$-axis represents the inclined plane (that is displacement along the inclined plane) and so the vertical represents the displacement vertical to this plane.
$$\begin{pmatrix} X_\text{inclined} \\ Y_\text{inclined}\end{pmatrix}=\begin{pmatrix} \cos(\phi) & \sin(\phi)\\ -\sin(\phi) & \cos(\phi) \end{pmatrix} \begin{pmatrix} X \\ Y\end{pmatrix}$$ In your case, $$\phi=25^o$$.