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:

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.

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

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