## Projectile Motion Formulae Derivation

There are a few general results that you may be asked to derive, but should not memorise for the A-Level exam. For a particle projected with initial velocity $$U$$ (note this is the actual velocity of the particle, not one of the components of velocity, hence the capital $$U$$) at an angle $$\alpha$$ above the horizontal and moving freely under gravity:

$$①$$ Time of flight =frac{2U sin alpha}{g}

$$②$$ Time to reach greatest height =frac{U sin alpha}{g}

$$③$$ Range on horizontal plane =frac{U^2 sin 2 alpha}{g}

$$④$$ Equation of trajectory: y=x tan alpha - frac{gx^2}{2U^2} (1+tan^2 alpha)

where y is the vertical height of the particle and x is the horizontal distance. In general, the path of such a projectile looks like so:

And I will refer to the following table in the derivations:

 $$\uparrow$$ $$\to$$ $$s=?$$ $$s=?$$ $$u=U\sin \alpha$$ $$v=U\cos \alpha$$ $$v=-U\sin \alpha$$  (by symmetry) $$t=?$$ $$a=-g$$ $$t=?$$

Where $$s$$ is displacement, $$u$$ is the initial velocity, $$v$$ is the final velocity, $$a$$ is the acceleration (and $$g \approx 9.8$$ is the acceleration due to gravity) and $$t$$ is the time.

$$\uparrow$$ represents the vertical component upwards and $$\to$$ represents the horizontal component to the right.

## Derivation of $$①:$$ Time of Flight

Using the vertical components:

$v=u+at$

$-U\sin \alpha=U\sin \alpha-gt$

$-2U\sin \alpha=-gt$

$\therefore t=\frac{2U \sin \alpha}{g}$

$Q.E.D.$

## Derivation of $$②:$$ Time to Reach Greatest Height

Using the vertical components, at the greatest height $$v=0$$, therefore we get:

$v=u+at$

$0=U\sin \alpha-gt$

$\therefore t=\frac{U \sin \alpha}{g}$

$Q.E.D.$

## Derivation of $$③:$$ Range on the Horizontal Plane

Using the same derivation as above with the vertical components, we get that $t=\frac{2U \sin \alpha}{g}$ Therefore, using v=frac{s}{t} :

$U\cos \alpha=\frac{s}{\frac{2U \sin \alpha}{g}}$

$U\cos \alpha \times \frac{2U \sin \alpha}{g}=s$

$s=\frac{2U^2 \cos \alpha \sin \alpha}{g}$

$s=\frac{U^2 (2 \sin \alpha \cos \alpha )}{g}$

$\therefore s=\frac{U^2 \sin 2 \alpha}{g}$

$Q.E.D.$

## Derivation of $$④:$$ The Equation of the Trajectory

Using v=frac{s}{t} with horizontal components gives:

$U\cos \alpha = \frac{x}{t}$

$\hspace{1cm} x=U\cos \alpha \times t \hspace{1cm} (\ast)$

Then using s=ut+frac{1}{2} a t^2 with vertical components gives:

$\hspace{1cm} y=U \sin \alpha \times t -\frac{1}{2} g t^2 \hspace{1cm} (\ast \ast)$

Rearranging $$(\ast)$$ for $$t$$ gives:

$\hspace{1cm} t=\frac{x}{U \cos \alpha} \hspace{1cm}$

Substituting this into $$(\ast \ast)$$ :

$y=U \sin \alpha \times \left(\frac{x}{U \cos \alpha}\right) -\frac{1}{2} g \left(\frac{x}{U \cos \alpha}\right)^2$

$y=x \tan \alpha -\frac{gx^2}{2U^2} \hspace{0.5mm} \sec^2 \alpha$

$y=x \tan \alpha -\frac{gx^2}{2U^2} (1+ \tan^2 \alpha)$

$Q.E.D.$