# A projectile is shot at an angle of pi/3  and a velocity of  11 m/s. How far away will the projectile land?

Nov 5, 2017

It will land 10.7 m from its point of launch.

#### Explanation:

If you have the benefit of the following formula, the result is quickly obtained:

$R = \frac{{v}_{o}^{2}}{g} \sin 2 \theta$

$R = \frac{{11}^{2}}{19.6} \sin \left(\frac{\pi}{3}\right) = \frac{121}{9.8} \left(0.866\right) = 10.7 m$

If you have not seen the above formula, it can be obtained through a (quite lengthy) calculation in which you use the equations of motion for a projectile, set up two equations in two unknowns, so that the time, $t$ can be eliminated and solve for the distance $x$ that the projectile moves horizontally. Many standard texts will show how this is done.

Wikipedia has the derivation here:

https://en.wikipedia.org/wiki/Range_of_a_projectile