Projectile Motion

Equation of motion

$$y=u_{y}t-\frac{1}{2}g{t}^2$$ $$x=u_{x}t$$ where $u_x=u\cos \theta$ and $u_y=u\sin \theta$.

Path (Trajectory)

Substituting $t=x/u_x$ into $y$, we see that $$y=\tan \theta \cdot x-\frac{g}{2u_x^2}\cdot x^2$$ No need to remember this equation.
But notice its form! $$y=\boxed{{\displaystyle \cdots }}x-\boxed{{\displaystyle \cdots }}{x}^2$$

Features of the form

1. Parabola ( $x^2$ )

2. First $\boxed{{\displaystyle \cdots }}$ describes the effect of projection angle
(or the slope of the projection line if no g)

3. Second $\boxed{{\displaystyle \cdots }}$ describes the bending effect of vertical acceleration g

4. Minus (-) sign: dragged down by gravity

5. Dimension of the second $\boxed{{\displaystyle \cdots }}$ must be 1/length, so that all terms have the same unit. In other words, [$u_x^2/g$] is some characteristic length.

6. The equation can also be written in a factorized form $$y=\text{scaling factor}\times x(R-x)$$ where $R$ is the range, the max. $x$.

Characteristic lengths and time

Half time of flight (time reaching the highest point): $${\tau }_{\ast }=\frac{u_y}{g}$$ Max. height (max. y): $$H=\frac{u_y^2}{2g}=\frac{1}{2}{u}_y{\tau }_{\ast }$$ Range (max. x): $$R=2u_x{\tau }_{\ast }=\frac{2u_x{u}_y}{g}$$ In terms of $R$, we can rewrite the trajectory as $$y=\frac{\tan \theta }{R}\cdot x(R-x)$$ Noting $R\tan \theta =4H$, we get $$y=4H\cdot \frac{x}{R}(1-\frac{x}{R})$$ which has the form $y=A{x}^{\prime }(1-x^{\prime })$.

Natural scales of the system

Length: $L=u^2/g$

Time: $T=u/g$

Dimensionless form and dependency of $\theta$

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