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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.

solving for vertical displacement and graph

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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.

You can go from one coordinate system to another by making rotations as follows:

$$\begin{pmatrix} X_\text{inclined} \\ Y_\text{inclined}\end{pmatrix}=\begin{pmatrix} \cos(\phi) & \sin(\phi)\\ -\sin(\phi) & \cos(\phi) \end{pmatrix} \begin{pmatrix} X \\ Y\end{pmatrix}$$ In your case, $\phi=25^o$.

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