# A projectile is shot at a velocity of 1 5 m/s and an angle of pi/8 . What is the projectile's peak height?

Jul 4, 2017

The peak height is $= 1.68 m$

#### Explanation:

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

The initial velocity is ${u}_{y} = v \sin \theta = 15 \cdot \sin \left(\frac{1}{8} \pi\right)$

The acceleration is $a = - g$

At the maximum height, $v = 0$

We apply the equation of motion

${v}^{2} = {u}^{2} + 2 a s$

to calculate the greatest height

$0 = \left(15 \sin {\left(\frac{1}{8} \pi\right)}^{2}\right) - 2 g h$

The greatest height is

$h = {\left(15 \sin \left(\frac{1}{8} \pi\right)\right)}^{2} / \left(2 g\right) = 1.68 m$