# A projectile is shot at an angle of pi/12  and a velocity of  6 m/s. How far away will the projectile land?

Jan 14, 2016

$1.84 \text{m}$

#### Explanation:

We treat the horizontal and vertical components of motion separately and use the fact that they share the same time of flight.

For the vertical component:

$v = u + a t$

This becomes:

${v}_{y} = 0 - \text{g} t$

Since ${v}_{y} = v \sin \theta$ this becomes:

$v \sin \theta = 0 - \text{g} t$

$\therefore t = \frac{v \sin \theta}{g}$

This is the time to reach maximum height so total time of flight $\Rightarrow$

t_("tot")=(2vsintheta)/g" "color(red)((1))

For the horizontal component:

$v \cos \theta = \frac{d}{t} _ \text{tot}$

$\therefore {t}_{\text{tot"=d/(vcostheta)" }} \textcolor{red}{\left(2\right)}$

Putting $\textcolor{red}{\left(1\right)}$ equal to $\textcolor{red}{\left(2\right)} \Rightarrow$

$\frac{2 v \sin \theta}{g} = \frac{d}{v \cos \theta}$

$\therefore d = \frac{{v}^{2} 2 \sin \theta \cos \theta}{g}$

Since $\sin 2 \theta = 2 \sin \theta \cos \theta$ this becomes:

$d = \frac{{v}^{2} \sin 2 \theta}{g} \text{ } \textcolor{red}{\left(3\right)}$

$\pi = {180}^{\circ}$
$\therefore \frac{\pi}{12} = \frac{180}{12} = {15}^{\circ} = \theta$
$\therefore 2 \theta = {30}^{\circ}$
$\therefore d = \frac{{6}^{2} \sin 30}{9.8} = \frac{36 \times 0.5}{9.8} = 1.84 \text{m}$
In future it is quicker if you can just remember equation $\textcolor{red}{\left(3\right)}$.