# A projectile is shot from the ground at an angle of (5 pi)/12  and a speed of 1 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 21, 2017

The distance is $= 0.054 m$

#### Explanation:

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

initial velocity is ${u}_{y} = v \sin \theta = 1 \cdot \sin \left(\frac{5}{12} \pi\right)$

Acceleration is $a = - g$

At the maximum height, $v = 0$

We apply the equation of motion

$v = u + a t$

to calculate the time to reach the greatest height

$0 = 1 \sin \left(\frac{5}{12} \pi\right) - g \cdot t$

$t = \frac{1}{g} \cdot \sin \left(\frac{5}{12} \pi\right)$

$= 0.099 s$

The greatest height is

$h = {\left(1 \sin \left(\frac{5}{12} \pi\right)\right)}^{2} / \left(2 g\right) = 0.048 m$

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

To find the horizontal distance, we apply the equation of motion

$s = {u}_{x} \cdot t$

$= 1 \cos \left(\frac{5}{12} \pi\right) \cdot 0.099$

$= 0.026 m$

The distance from the starting point is

$d = \sqrt{{h}^{2} + {s}^{2}}$

$= \sqrt{{0.048}^{2} + {0.026}^{2}}$

$= 0.054 m$