# If a projectile is shot at an angle of pi/3 and at a velocity of 32 m/s, when will it reach its maximum height??

at time $t = 2.827838053 \text{ }$seconds

#### Explanation:

The equation for the vertical displacement for projectiles

$y = {v}_{0} \sin \theta \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$

The equation for the horizontal displacement for projectiles is

$x = {v}_{0} \cos \theta \cdot t$
Maximizing the vertical displacement by obtaining the first derivative of y with respect to time t

$\frac{\mathrm{dy}}{\mathrm{dt}} = {v}_{0} \sin \theta + \frac{1}{2} \cdot g \cdot 2 \cdot t$

set $\frac{\mathrm{dy}}{\mathrm{dt}} = 0$

${v}_{0} \sin \theta + \frac{1}{2} \cdot g \cdot 2 \cdot t = 0$

$t = \frac{- {v}_{0} \cdot \sin \theta}{g}$

$t = \frac{- 32 \sin \left(\frac{\pi}{3}\right)}{- 9.8}$

$t = \frac{80 \sqrt{3}}{49}$

$t = 2.827838053 \text{ }$seconds

God bless....I hope the explanations is useful.