# A projectile is shot at a velocity of  1 m/s and an angle of pi/4 . What is the projectile's peak height?

Mar 10, 2018

If the projectile is shot at a velocity of $1$ m/s then the vertical component of velocity is given by $1 \cdot \sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ m/s and the horizontal component of velocity is given by $1 \cdot \cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ m/s.

The maximum height will occur when the vertical component of velocity equals $0$.

We now need to find a kinematic equation that will make sense for this problem.

${v}_{y}^{2} = {v}_{{y}_{0}}^{2} + 2 {a}_{y} \left(\Delta y\right)$

We will be solving for $\Delta y$, the maximum height of the projectile. ${a}_{y}$ will be the acceleration due to gravity, $9.8 \text{ } \frac{m}{s} ^ 2$.

$0 = \frac{1}{2} + 2 \left(- 9.8\right) \Delta y$

$\Delta y = 0.026$ meters, or $2.6$ cm.

Hopefully this helps!