# A projectile is shot at an angle of pi/3  and a velocity of  21 m/s. How far away will the projectile land?

Mar 22, 2017

The distance is $= 39 m$

#### Explanation:

Solving in the vertical direction ${\uparrow}^{+}$

${v}_{y} = u \sin \theta = 21 \cdot \sin \left(\frac{\pi}{3}\right)$

We apply the equation

$s = u t + \frac{1}{2} a {t}^{2}$

To find the time

$s = 0$

$0 = t \cdot 21 \sin \left(\frac{\pi}{3}\right) - \frac{1}{2} \cdot 9.8 \cdot {t}^{2}$

$t \left(21 \sin \left(\frac{\pi}{3}\right) - 4.9 t\right) = 0$

Therefore,

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

and

$t = \frac{21}{4.9} \sin \left(\frac{\pi}{3}\right)$ when the projectile will land

Solving in the horizontal direction, ${\to}^{+}$

${v}_{x} = u \cos \theta = 21 \cos \left(\frac{\pi}{3}\right)$

Distance is

d=21tcospi/3)=21*21/4.9*sin(pi/3)cos(pi/3)

$= 39 m$