# A projectile is shot from the ground at an angle of pi/8  and a speed of 5 /8 m/s. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Nov 18, 2016

The distance is $= 0.014 m$

#### Explanation:

The maximum height is when the vertical component of the velocity $= 0$

We use the equation,

$v = u - g \cdot t$

$v = 0$

$u = {u}_{0} \sin \theta = \frac{5}{8} \cdot \sin \left(\frac{\pi}{8}\right)$

$0 = \frac{5}{8} \sin \left(\frac{\pi}{8}\right) - g \cdot t$

$t = \frac{5 \sin \left(\frac{\pi}{8}\right)}{\left(8 g\right)}$

This is the time to reach the maximum height.

To, get the horizontal distance from the starting point, we multiply by the horizontal component of the velocity.

$x = {u}_{0} \cos \theta \cdot t$

$x = \frac{5}{8} \cdot \cos \left(\frac{\pi}{8}\right) \cdot 5 \sin \frac{\frac{\pi}{8}}{8 g}$

$x = \left(\frac{25}{64 g}\right) \cdot \frac{1}{2} \cdot \sin \left(\frac{\pi}{4}\right)$

$x = \frac{25 \sqrt{2}}{256 g} = 0.014 m$