# A projectile is shot at a velocity of 2 m/s and an angle of pi/8 . What is the projectile's peak height?

Dec 29, 2015

The projectile's peak height is $23.69 m$

#### Explanation:

You have to consider the two component of motion
$s \left(x\right) = v t$ with v=vcos22.5°
$s \left(y\right) = v t - \frac{1}{2} a {t}^{2}$ with v=vsin22.5°
You have to find the moment when the vertical speed is =0, that is the moment of the peak.
So $v = v - a t = 0$ , $t = \frac{g}{v} = 12.82 s$
Use this value in the equation of horizontal component
s(x)=(vcos22.5°)*12.82=23.69m