A projectile is shot from the ground at an angle of (5 pi)/12  and a speed of 6/5 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 19, 2017

$\mathrm{dx} \approx 0.03658 m , \mathrm{dy} \approx 0.205 m$

Explanation:

I copied, pasted and changed the values of this answer:
https://socratic.org/questions/a-projectile-is-shot-from-the-ground-at-an-angle-of-pi-6-and-a-speed-of-5-m-s-wh#425323
If you have difficulties understanding projectiles motion visit this:
https://socratic.org/physics/2d-motion/projectile-motion

We have to calculate the initial speed in x and in y:
${V}_{0 x} = {V}_{o} \cos \left(\theta\right) = \left(\frac{6}{5}\right) \cos \left(\frac{5 \pi}{12}\right) = 0.31 \frac{m}{s}$
${V}_{0 y} = {V}_{o} \sin \left(\theta\right) = \left(\frac{6}{5}\right) \sin \left(\frac{5 \pi}{12}\right) = 1.159 \frac{m}{s}$

Let's calculate the distance in y (height):

We have to know when is the projectile at its maximum height. Let's use the equation of the speed for an object with acceleration:
${V}_{y} = {V}_{0 y} + a \Delta t$

The acceleration is $- 9.8 \frac{m}{s} ^ 2$ because of earth's gravity:
$0 = 1.159 + \left(- 9.8\right) \Delta t \implies \Delta t \approx 0.118 s$

To find the distance in y we have to use the equation of distance:
$y = {y}_{0} + {v}_{0 y} t + \frac{1}{2} a {t}^{2}$
$\implies y = \left(1.159\right) \left(0.118\right) + \left(\frac{1}{2}\right) \left(- 9.8\right) {\left(0.118\right)}^{2}$
$\implies y \approx 0.205 m$

To find the distance in x we have to use another time the equation of distance (this time there's no acceleration):
$x = {x}_{0} + {v}_{0 x} t + \frac{1}{2} a {t}^{2}$
$\implies x = \left(0.31\right) \left(0.118\right)$
$\implies x \approx 0.03658 m$