# Projectile motion question: vertical and horizontal graph

I've have got some vertical and horizontal distances for a projectile-like motion.

In order to work out the trajectory, why is it better to plot on the x-axis, "horizontal distance^2", and on the y axis, "vertical distance"?

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A trajectory equation should have the same units on both axis. You simply plot the position along X and position along y. – Antillar Maximus Nov 6 '11 at 19:20
Okay, so i plooted the vertical and horizontal results. I got a weird graph, see here link. The black line is the line of best fit, and the blue is the original. What can i say about the trajectory from this graph? – fas Nov 6 '11 at 20:12
All you can say is that the trajectory is linear. In regular projectile motion, if x is very small, you can neglect the x^2 term and obtain a linear relationship between y and x. I am using: y=xtan(th)-gx^2/2v^2cos^2(th). Not sure if I answered your question. Perhaps you can give us some more information? – Antillar Maximus Nov 6 '11 at 21:26

## 1 Answer

The equations of motion, given the usual x and y axis choices, are $y=-0.5gt^2 +v_yt +y_0$ and $x=v_xt + x_0$. That is, the initial conditions are velocity $v_x$ or $v_y$, and position $y_0$ or $x_0$. The only difference is that gravity operates in the y direction.

Plotting y versus x gives a parabola.

Plotting y versus $x^2$ means you are plotting points of the form:
$(y,x) = (-0.5gt^2 + v_yt + y_0, v_x^2t^2 + 2v_xx_0t + x_0^2)$
For large $t$, these points will approach a straight line through the origin with slope $-0.5g/v_x^2$.

Not sure how this gives a great advantage in plotting. Maybe you are to assume that $x(0) = y(0) = 0$, which makes it a little simpler. Then you can get $v_y$ from the behavior near $(0,0)$.

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