# A projectile is shot from the ground at an angle of pi/6  and a speed of 4 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?

Nov 25, 2016

The maximum height is $= 0.2 m$

The horizontal distance is $= 0.69 m$

#### Explanation:

The maximum height is when the vertical component of the velocity $= 0$

We use the equation

${v}^{2} = {u}^{2} - 2 g y$

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

$y = {\left(4 \cdot \sin \left(\frac{\pi}{6}\right)\right)}^{2} / \left(2 g\right)$

${y}_{\max} = \frac{2}{g} = \frac{2}{10} = 0.2 m$

The time to reach the greatest height is $t = {v}_{0} \sin \frac{\theta}{g}$

$t = 4 \cdot \sin \frac{\frac{\pi}{6}}{10} = 0.2 s$

The horizontal distance is $x = {v}_{0} \cos \theta \cdot t$

$= 4 \cdot \cos \left(\frac{\pi}{6}\right) \cdot 0.2 = 0.8 \cdot \frac{\sqrt{3}}{2} = 0.4 \sqrt{3} = 0.69 m$

graph{y=(-5x^2/12)+0.58x [-0.819, 1.881, -0.032, 1.318]}