# A projectile is shot from the ground at an angle of pi/8  and a speed of 3 m/s. When the projectile is at its maximum height, what will its distance from the starting point be?

May 18, 2017

The distance is $= 0.12 m$

#### Explanation:

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

initial velocity is ${u}_{y} = v \sin \theta = 3 \cdot \sin \left(\frac{1}{8} \pi\right)$

Acceleration is $a = - g$

At the maximum height, $v = 0$

We apply the equation of motion

$v = u + a t$

to calculate the time to reach the greatest height

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

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

$= 0.12 s$

The greatest height is

$h = {\left(3 \sin \left(\frac{\pi}{8}\right)\right)}^{2} / \left(2 g\right) = 0.067 m$

Resolving in the horizontal direction ${\rightarrow}^{+}$

To find the horizontal distance, we apply the equation of motion

$s = {u}_{x} \cdot t$

$= 3 \cos \left(\frac{1}{8} \pi\right) \cdot 0.12$

$= 0.1 m$

The distance from the starting point is

$d = \sqrt{{h}^{2} + {s}^{2}}$

$= \sqrt{{0.07}^{2} + {0.1}^{2}}$

$= 0.12 m$