# A projectile is shot from the ground at an angle of pi/4  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?

$d = 8.2141272642757 \text{ }$meters

#### Explanation:

Given $\theta = \frac{\pi}{4}$
initial velocity ${v}_{0} = 12 \text{ }$m/sec

Compute the total time t it will be on the air

$y = {v}_{0} \cdot \sin \theta \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$

$0 = 12 \cdot \sin \left(\frac{\pi}{4}\right) \cdot t + \frac{1}{2} \left(- 9.8\right) \cdot {t}^{2}$

$0 = 12 \cdot \left(\frac{1}{\sqrt{2}}\right) \cdot t - 4.9 \cdot {t}^{2}$

$0 = 6 \sqrt{2} \cdot t - 4.9 \cdot {t}^{2}$

$0 = t \left(6 \sqrt{2} - 4.9 t\right)$

${t}_{1} = 0$ and ${t}_{2} = \frac{6 \sqrt{2}}{4.9}$

Let $x =$range

We will use ${t}_{2} = \frac{6 \sqrt{2}}{4.9}$

$x = {v}_{0} \cos \theta \cdot t$

$x = 12 \cdot \cos \left(\frac{\pi}{4}\right) \frac{6 \sqrt{2}}{4.9}$

$x = 14.69387755 \text{ }$meters

We need the half-range $\frac{x}{2} = 7.3469387755 \text{ }$meters

We also need the maximum height ${y}_{\max}$

$y = {v}_{0} \cdot \sin \theta t + \frac{1}{2} g \cdot {t}^{2}$

At ${t}_{2} / 2$ the projectile is at its maximum height.

${y}_{\max} = {v}_{0} \cdot \sin \theta \cdot {t}_{2} / 2 + \frac{1}{2} g \cdot {\left({t}_{2} / 2\right)}^{2}$

${y}_{\max} = 12 \sin \left(\frac{\pi}{4}\right) \frac{\frac{6 \sqrt{2}}{4.9}}{2} + \frac{1}{2} \left(- 9.8\right) {\left(\frac{\frac{6 \sqrt{2}}{4.9}}{2}\right)}^{2}$

${y}_{\max} = 3.673469387755 \text{ }$meters

Distance $d$ of the projectile from the starting point

$d = \sqrt{{\left(\frac{x}{2}\right)}^{2} + {y}_{\max}^{2}}$

$d = \sqrt{{\left(7.3469387755\right)}^{2} + {3.673469387755}^{2}}$

$d = 8.2141272642757 \text{ }$meters

God bless....I hope the explanation is useful.