# A projectile is shot from the ground at an angle of pi/6  and a speed of 3 m/s. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?

##### 1 Answer

${y}_{\max} = + 0.114795 \text{ }$meter$= 11.4795 \text{ }$centimeter
$x = + 0.3976647262 \text{ }$meter$= 39.76647262 \text{ }$centimeter

#### Explanation:

Equations for projectiles are as follows

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

Time from the ground to its maximum height
$t = \frac{- {v}_{0} \sin \theta}{g}$

Equation for the maximum height ${y}_{\max}$

${y}_{\max} = {v}_{0} \sin \theta \cdot \frac{- {v}_{0} \sin \theta}{g} + \frac{1}{2} \cdot g {\left(\frac{- {v}_{0} \sin \theta}{g}\right)}^{2}$

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

${y}_{\max} = {v}_{0}^{2} \cdot {\sin}^{2} \theta \left(\frac{1}{2 g} - \frac{1}{g}\right)$

${y}_{\max} = \frac{- {v}_{0}^{2} \cdot {\sin}^{2} \theta}{2 g}$

${y}_{\max} = \frac{- {\left(3\right)}^{2} \cdot {\left(\sin {30}^{\circ}\right)}^{2}}{2 \cdot \left(- 9.8\right)}$

${y}_{\max} = + 0.114795 \text{ }$ meter

Horizontal distance

$x = {v}_{0} \cdot \cos \theta \cdot t$

$x = 3 \cdot \cos {30}^{\circ} \left(\frac{- {v}_{0} \sin \theta}{g}\right)$

$x = 3 \cdot \cos {30}^{\circ} \cdot \left(\frac{- 3 \cdot \sin \left({30}^{\circ}\right)}{- 9.8}\right)$

$x = + 0.3976647262 \text{ }$meter

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