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.