# A projectile is shot from the ground at an angle of pi/8  and a speed of 12 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?

Feb 2, 2017

The horizontal distance is $= 5.2 m$

#### Explanation:

Solving in the vertical direction ${\uparrow}^{+}$

$u = 12 \sin \left(\frac{\pi}{8}\right) m {s}^{-} 1$

$v = 0 m {s}^{-} 1$

$a = - g m {s}^{-} 2$

$v = u + a t$

$0 = 12 \sin \left(\frac{\pi}{8}\right) - g t$

$t = \frac{12 \sin \left(\frac{\pi}{8}\right)}{g}$

This is the time to reach the greatest height

Solving in the horizontal direction ${\rightarrow}^{+}$

$u = 12 \cos \left(\frac{\pi}{8}\right)$

$s = u \cdot t = \frac{12 \sin \left(\frac{\pi}{8}\right)}{g} \cdot 12 \cos \left(\frac{\pi}{8}\right)$

$= \frac{144}{g} \cdot \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right)$

$= 5.2 m$