# If a projectile is shot at an angle of (7pi)/12 and at a velocity of 2 m/s, when will it reach its maximum height?

Apr 23, 2016

#### Answer:

Time $t = \frac{5 \sqrt{6} + 5 \sqrt{2}}{98} = 0.1971277197 \text{ }$second

#### Explanation:

For the vertical displacement $y$

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

We maximize displacement $y$ with respect to $t$

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

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

set $\frac{\mathrm{dy}}{\mathrm{dt}} = 0$ then solve for $t$

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

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

$t = \frac{- 2 \cdot \sin \left(\frac{7 \pi}{12}\right)}{- 9.8}$

Note: $\sin \left(\frac{7 \pi}{12}\right) = \sin \left(\frac{5 \pi}{12}\right) = \frac{\sqrt{6} + \sqrt{2}}{4}$

$t = \frac{- 2 \cdot \frac{\left(\sqrt{6} + \sqrt{2}\right)}{4}}{- 9.8}$

$t = \frac{5 \sqrt{6} + 5 \sqrt{2}}{98} = 0.1971277197 \text{ }$second

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