# A projectile is shot from the ground at an angle of pi/12  and a speed of 2 /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?

Nov 9, 2017

Split this into two parts, horizontal and vertical

#### Explanation:

The initial velocity will be split into both horizontal and vertical components. For horizontal, ${v}_{0 x} = {v}_{0} C o s \left(\theta\right)$ and for vertical it's ${v}_{0 y} = {v}_{0} S \in \left(\theta\right)$

So, for you ${v}_{0 x} = \frac{2}{3} C o s \left(\frac{\pi}{12}\right) \mathmr{and} {v}_{0} y = \frac{2}{3} S \in \left(\frac{\pi}{12}\right)$ Be sure your calculator is in radian mode.

Then, as soon as it launches, gravity will take over, so there will be an acceleration of $- 9.8 \frac{m}{s} ^ 2$. Keys:

$1 t h e r e i s n o a \mathcal{e} \le r a t i o n \in t h e x \mathrm{di} r e c t i o n \mathmr{and}$v_f = v_0 absent of friction or air resistance.

$2 , \in t h e y \mathrm{di} r e c t i o n ,$v_{fy}=0 because the projectile will come to a momentary stop.

use equations $d = {v}_{0} t + \frac{1}{2} a {t}^{2} \mathmr{and} {v}_{f} = {v}_{0} + a t$ as needed for each the x and y components using what you know. The first thing you need to find is the time, noting that the time for x and y is the same because the projectile hits the ground once.

For y: ${v}_{0 y} = \frac{2}{3} \frac{m}{s} S \in \left(\frac{\pi}{12}\right)$
${v}_{f y} = 0$
${a}_{y} = - 9.8 \frac{m}{s} ^ 2$
t=?
d=?

For x: ${v}_{0 x} = \frac{2}{3} \frac{m}{s} C o s \left(\frac{\pi}{12}\right)$
${v}_{f x} = {v}_{0 x}$
${a}_{x} = 0$
t= the same time for y
d=?