# A projectile is shot at an angle of (5pi)/12  and a velocity of  8 m/s. How far away will the projectile land?

Nov 13, 2016

Distance $= 3.2 m$

#### Explanation:

Initial velocity $u = 8 m {s}^{- 1}$ at $\theta = \frac{5 \pi}{12}$

We are going to use 2 equations

$h = u t \sin \theta - g {t}^{2} / 2$

Where $h =$ height

Here $h = 0$

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

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

The second equation is

$d = u t \cos \theta$

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

$\sin \left(2 \theta\right) = 2 \sin \theta \cos \theta$

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

$\sin \left(5 \frac{\pi}{6}\right) = \frac{1}{2}$

Therefore $d = {8}^{2} \sin \frac{\frac{5 \pi}{6}}{g} = 64 \cdot \frac{1}{2} \cdot \frac{1}{g}$

$= \frac{32}{10} = 3.2 m$

graph{3.73x-1.17x^2 [-2.06, 4.098, -0.144, 2.936]}