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

May 15, 2016

Resolve velocity into x and y components:
${v}_{x} = 12 \cos \left(\frac{\pi}{3}\right) = 6.00 m {s}^{-} 1$
${v}_{y} = 12 \sin \left(\frac{\pi}{3}\right) = 10.4 m {s}^{-} 1$

Maximum height $\implies {v}_{y} = 0$

Constant acceleration, given $v$ and $a$ so use this kinematic equation:
${v}_{y}^{2} - {v}_{y 0}^{2} = 2 {a}_{y} \left({s}_{y} - {s}_{y 0}\right)$
Solve for ${s}_{y}$:
$0 - {10.4}^{2} = 2 \cdot - 9.81 \cdot {s}_{y} \implies {s}_{y} = 5.50 m$

Solve for $t$:
${v}_{y} - {v}_{y 0} = {a}_{y} \left(t - {t}_{0}\right) \implies t = 1.06 s$
Air resistance assumed negligible $\implies {a}_{x} = 0$
Solve for ${s}_{x}$:
${s}_{x} - {s}_{x 0} = {v}_{x 0} \left(t - {t}_{0}\right) + \frac{1}{2} {a}_{x} {\left(t - {t}_{0}\right)}^{2}$
${s}_{x} = 6 \cdot 1.06 = 6.36 m$

Total distance travelled = $\sqrt{{s}_{x}^{2} + {s}_{y}^{2}} = 8.41 m$