# A projectile is shot at an angle of pi/4  and a velocity of  5 m/s. How far away will the projectile land?

Jan 18, 2016

If the projectile starts and ends at the same height (ie ground level), we can use the range formula, $\Delta {d}_{x} = \frac{{v}_{i}^{2} \cdot \sin \left(2 \theta\right)}{g}$

#### Explanation:

A word of caution - the range formula is not universal. It is a derived formula using $\Delta {d}_{y} = 0$, in other words the start and end height are the same. If they are not the same (eg the projectile lands on a roof), a different technique must be used.

The range formula is

$\Delta {d}_{x} = \frac{{v}_{i}^{2} \cdot \sin \left(2 \theta\right)}{g}$

where
$\Delta {d}_{x}$ is the horizontal distance (range)
${v}_{i}$ is the magnitude of the launch velocity
$\theta$ is the launch angle
$g$ is the magnitude of the acceleration of gravity (usually 9.8, 9.81, or 10, depending on how the course is presented).

In this case,

$\Delta {d}_{x} = \frac{{5}^{2} \cdot \sin \left(2 \frac{\pi}{4}\right)}{g}$