# A projectile is shot at an angle of pi/8  and a velocity of  4 "m/s". How far away will the projectile land?

Dec 10, 2016

The distance is $= 1.15 m$

#### Explanation:

To find the range of the projectile, the height $h = 0$

We use the equation

$h = {u}_{0} t \sin \theta - \frac{1}{2} g {t}^{2}$

$t \left({u}_{0} \sin \theta - \frac{1}{2} g t\right) = 0$

$t = 0$ is the time when the projectile is shot

$t = \frac{2 {u}_{0} \sin \theta}{g}$

The horizontal component of the velocity $= {u}_{0} \cos \theta$

So, the range

$= \frac{2 {u}_{0} \sin \theta}{g} \cdot {u}_{0} \cos \theta$

$= {u}_{0}^{2} \sin \frac{2 \theta}{g}$

as, $2 \sin \theta \cos \theta = \sin 2 \theta$

The range

$= {u}_{0}^{2} \sin \frac{2 \theta}{g} = {4}^{2} \cdot \sin \left(\frac{\pi}{4}\right) \cdot \frac{1}{9.8}$

$= 16 \cdot \frac{\sqrt{2}}{2} \cdot \frac{1}{9.8} = 1.15 m$