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) \(t=?\)

 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:


\[-U\sin \alpha=U\sin \alpha-gt\]

\[-2U\sin \alpha=-gt\]

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



Derivation of \(②:\) Time to Reach Greatest Height

Using the vertical components, at the greatest height \(v=0\), therefore we get:


\[0=U\sin \alpha-gt\]

\[\therefore t=\frac{U \sin \alpha}{g}\]



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}` :

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

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

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

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

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



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 \((\ast \ast)\) :

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

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

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



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