# A projectile is shot from the ground at an angle of pi/4  and a speed of 3/2 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 31, 2017

The distance is $= 0.13 m$

#### Explanation:

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

initial velocity is ${u}_{y} = v \sin \theta = \frac{3}{2} \cdot \sin \left(\frac{1}{4} \pi\right)$

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

$0 = \frac{3}{2} \sin \left(\frac{1}{4} \pi\right) - g \cdot t$

$t = \frac{3}{2 g} \cdot \sin \left(\frac{1}{4} \pi\right)$

$= 0.108 s$

The greatest height is

$h = {\left(\frac{3}{2} \sin \left(\frac{1}{4} \pi\right)\right)}^{2} / \left(2 g\right) = 0.057 m$

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

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

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

$= \frac{3}{2} \cos \left(\frac{1}{4} \pi\right) \cdot 0.108$

$= 0.115 m$

The distance from the starting point is

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

$= \sqrt{{0.057}^{2} + {0.115}^{2}}$

$= 0.13 m$