# A projectile is shot from the ground at an angle of ( pi)/3  and a speed of 7/4 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?

Feb 1, 2017

Horizontal distance $= 0.135 m$

#### Explanation:

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

$u = \frac{7}{4} \sin \left(\frac{\pi}{3}\right) m {s}^{-} 1$

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

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

$v = u + a t$

$0 = \frac{7}{4} \sin \left(\frac{\pi}{3}\right) - g t$

$t = \frac{\frac{7}{4} \sin \left(\frac{\pi}{3}\right)}{g}$

This is the time to reach the greatest height

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

$u = \frac{7}{4} \cos \left(\frac{\pi}{3}\right)$

$s = u \cdot t = \frac{\frac{7}{4} \sin \left(\frac{\pi}{3}\right)}{g} \cdot \frac{7}{4} \cos \left(\frac{\pi}{3}\right)$

$= \frac{49}{16 g} \cdot \sin \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}\right)$

$= 0.135 m$