# A projectile is shot from the ground at an angle of pi/8  and a speed of 19 m/s. When the projectile is at its maximum height, what will its distance from the starting point be?

Apr 12, 2017

The distance is $= 13.02 m$

#### Explanation:

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

initial velocity is ${u}_{y} = v \sin \theta = 19 \cdot \sin \left(\frac{1}{8} \pi\right)$

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

$0 = 19 \sin \left(\frac{1}{8} \pi\right) - g \cdot t$

$t = 19 \cdot \frac{1}{g} \cdot \sin \left(\frac{1}{8} \pi\right)$

$= 0.74 s$

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

We apply the equation of motion

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

$= 19 \cos \left(\frac{1}{8} \pi\right) \cdot 0.74$

$= 13.02 m$