# A projectile is shot at a velocity of  23 m/s and an angle of pi/4 . What is the projectile's peak height?

Apr 9, 2018

To answer this the formulas ${u}_{y} = u \sin \left(\theta\right)$, ${v}_{y} = {u}_{y} + {a}_{y} t$ and $y = {u}_{y} t + \frac{{a}_{y} {t}^{2}}{2}$ can be used.

#### Explanation:

First we want vertical component of velocity
${u}_{y} = u \sin \left(\theta\right)$
${u}_{y} = 23 \sin \left(\frac{\pi}{4}\right)$
${u}_{y} = \frac{23}{\sqrt{2}}$

Second we find the time taken to reach a vertical velocity of $0$ (this is its peak height).
${v}_{y} = {u}_{y} + {a}_{y} t$
$0 = \frac{23}{\sqrt{2}} - \frac{9.8}{t}$
$t = 9.8 \times \frac{\sqrt{2}}{23}$
(Assuming the projectile is on Earth ${a}_{y} = - 9.8$ )

Finally we can find the peak height
$y = {u}_{y} t + \frac{{a}_{y} {t}^{2}}{2}$
$y = \frac{23}{\sqrt{2}} \times 9.8 \times \frac{\sqrt{2}}{23} - 9.8 \times {\left(9.8 \times \frac{\sqrt{2}}{23}\right)}^{2} / 2$

$y = 9.8 - {9.8}^{3} / {23}^{2}$
$y = 8.02$ $m$