# A projectile is shot from the ground at an angle of pi/6  and a speed of 12 m/s. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Nov 17, 2016

The distance is $d \approx 6.5 m$

#### Explanation:

Write the equations for y(t) and x(t):

$y \left(t\right) = \left(- 4.9 \frac{m}{s} ^ 2\right) {t}^{2} + \left(12 \frac{m}{s}\right) \sin \left(\frac{\pi}{6}\right) t$

$x \left(t\right) = \left(12 \frac{m}{s}\right) \cos \left(\frac{\pi}{6}\right) t$

Using the t coordinate for the axis of symmetry for y(t):

$h = - \frac{b}{2 a} = - \frac{\left(12 \frac{m}{s}\right) \sin \left(\frac{\pi}{6}\right)}{2 \left(- 4.9 \frac{m}{s} ^ 2\right)} \approx 0.6 s$

The distance, d, is:

$d = \sqrt{{\left(\left(- 4.9 \frac{m}{s} ^ 2\right) {\left(0.6 s\right)}^{2} + \left(12 \frac{m}{s}\right) \sin \left(\frac{\pi}{6}\right) 0.6 s\right)}^{2} + {\left(\left(12 \frac{m}{s}\right) \cos \left(\frac{\pi}{6}\right) 0.6 s\right)}^{2}}$

$d \approx 6.5 m$