# If a projectile is shot at a velocity of 15 m/s and an angle of pi/4, how far will the projectile travel before landing?

Nov 14, 2016

The distance is $= 22.5 m$

#### Explanation:

$u = 15 m {s}^{- 1}$

$\theta = \frac{\pi}{4}$

We use the equation $y \left(t\right) = u \sin \theta \cdot t - \frac{1}{2} \cdot g \cdot {t}^{2}$

to determine the time of flight.

$0 = u \sin \theta \cdot t - \frac{1}{2} g {t}^{2}$

$u \sin \theta = \frac{1}{2} \cdot g \cdot t$

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

The horizontal distance is $d = u \cos \theta \cdot \frac{2 u \sin \theta}{g}$

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

Let $g = 10 m {s}^{- 2}$

$d = {15}^{2} \cdot \sin \frac{\frac{\pi}{2}}{10} = 225 \cdot \frac{1}{10} = 22.5 m$

graph{x-(x^2/22.5) [-12.5, 38.8, -0.83, 24.84]}