# A projectile is shot at an angle of pi/4  and a velocity of  19 m/s. How far away will the projectile land?

Nov 6, 2016

The range is $= 36.1 m$

#### Explanation:

We use two formulas
$x = {v}_{0} \cos \theta \cdot t$
$h = {v}_{0} \sin \theta \cdot t - \frac{1}{2} g \cdot {t}^{2}$
Here we have ${v}_{0} = 19 \frac{m}{s}$ and $\theta = \frac{\pi}{4}$
$h = 0$
$0 = {v}_{0} \sin \theta \cdot t - \frac{1}{2} g \cdot {t}^{2}$
$\therefore t = 0$ or $t = 2 {v}_{0} \sin \frac{\theta}{g}$
So $x = {v}_{0}^{2} \cdot 2 \sin \theta \cdot \cos \frac{\theta}{g} = \frac{{v}_{0}^{2} \cdot \sin 2 \theta}{g}$
$x = 19 \cdot 19 \cdot \sin \left(\frac{\pi}{2}\right) \cdot \frac{1}{g} = \frac{361}{10} = 36.1 m$