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

Jan 11, 2017

The distance is $= 0.3975 m$

#### Explanation:

(1) The vertical motion ${\uparrow}^{+}$

$u = 3 \sin \left(\frac{\pi}{3}\right)$

$a = - 9.8$

$v = 0$

$s = h$

${v}^{2} = {u}^{2} + 2 a s$

$\implies$, $0 = {\left(3 \sin \left(\frac{\pi}{3}\right)\right)}^{2} - 2 \cdot 9.8 \cdot h$

$h = {\left(3 \sin \left(\frac{\pi}{3}\right)\right)}^{2} / \left(2 \cdot 9.8\right) = 0.3439 m$

(2) Time to reach greatest height

$v = u + a t$

$v = u + g t$

$0 = 3 \sin \left(\frac{\pi}{3}\right) - 9.8 \cdot t$

$t = \frac{3 \sin \left(\frac{\pi}{3}\right)}{9.8} = 0.265 s$

(3) The horozontal motion ${\rightarrow}^{+}$

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

$d = u t = 3 \cos \left(\frac{\pi}{3}\right) \cdot 0.265 = 0.3975 m$