Projectile motion formulae

In a nutshell

It is possible to derive formulae for the time of flight, the horizontal range and the time taken to reach the greatest height.

Formulae

Consider a particle projected at an angle on a horizontal plane:

Maths; Projectiles; KS5 Year 13; Projectile motion formulae

The following formulae can be derived:

Time of flight

Consider the vertical motion for the time the particle is in flight:

$u_y=U\sin\alpha \quad s=0 \quad a=-g \quad t=T$

Substitute into $s=ut+\dfrac{1}{2}at^2$ to give:

$\begin{aligned}s&=ut+\dfrac{1}{2}at^2\\\\0&=(U\sin\alpha)T-\dfrac{g}{2}T^2\\\\0&=T(U\sin\alpha-\dfrac{gT}{2})\\\\T&=0 \ or \ T =\dfrac{2U\sin\alpha}{g}\end{aligned}$

Therefore:

$\boxed{\text{Time of flight}=\cfrac{2U\sin\alpha}{g}}$

Horizontal range

Consider the horizontal motion:

$\begin{aligned} u_x&=U\cos\alpha \ \quad s=R \quad t=T\\\ \\s&=v t\\ R&=U\cos\alpha \times T \end{aligned}$

Substitute in the time of flight $T=\cfrac{2U\sin\alpha}{g}$

$\begin{aligned}R&=U\cos\alpha\times\dfrac{2U\sin\alpha}{g} \\\\&=\dfrac{U^2 \times 2\sin\alpha\cos\alpha}{g}\end{aligned}$

Substituting in using the double angle formula $2\sin\alpha\cos\alpha\equiv\sin2\alpha$ gives the horizontal range as:

$\boxed{\text{Range}=\cfrac{U^2\sin2\alpha}{g}}$

Greatest height

At the greatest height, the vertical component of the velocity is zero:

$u = U\sin \alpha \quad v=0 \quad a=-g \quad t=T$

Use $v=u+at$ to give:

$\begin{aligned}v &=u + at \\0 &= U \sin\alpha - gT\end{aligned}$

Rearrange for $T$ to get the time to reach the greatest height:

$\boxed{\text{Time to reach greatest height}=\cfrac{U\sin\alpha}{g}}$

Example

A particle is projected at $10\ m.s^{-1}$  at $30\degree$ above the horizontal. Find the time of flight and horizontal range.

$\begin{aligned}\text{Time of flight}&=\dfrac{2U\sin\alpha}{g} \\\\&=\dfrac{2\times10\times\sin30}{9.8} \\\\&=\underline{1.02\ s} \ (3\thinspace s.f.)\end{aligned}$ 

$\begin{aligned}\text{Range}&=\cfrac{U^2\sin2\alpha}{g} \\\\&=\cfrac{10^2\times\sin60}{9.8} \\\\&=\underline{8.84\ m} \ (3 \thinspace s.f.)\end{aligned}$ 