# A projectile is shot from the ground at an angle of pi/8  and a speed of 3 /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?

Mar 24, 2017

The distance is $= 0.005 m$

#### Explanation:

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

$u = {u}_{0} \sin \theta = \frac{3}{8} \sin \left(\frac{1}{8} \pi\right)$

$a = - g$

$v = 0$ at the maximum height

We apply the equation

$v = u + a t$

$t = \frac{v - u}{a} = \frac{0 - \frac{3}{8} \sin \left(\frac{1}{8} \pi\right)}{- g}$

$= \frac{3}{8 g} \sin \left(\frac{1}{8} \pi\right) = 0.0146 s$

Time to reach the maximum height is $= 0.0146 s$

Solving in the horizontal direction $\to +$

distance $d = {u}_{0} t \cos \left(\frac{1}{8} \pi\right) = 0.0146 \cdot \frac{3}{8} \cdot \cos \left(\frac{1}{8} \pi\right)$

$= 0.005 m$