# A projectile is shot from the ground at an angle of pi/6  and a speed of 5 m/s. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?

May 18, 2017

dx=0,22 m dy=0,9566

#### Explanation:

We have to calculate the initial speed in x and in y:
${V}_{0 x} = {V}_{o} \cos \left(\theta\right) = \left(5\right) \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$
${V}_{0 y} = {V}_{o} \sin \left(\theta\right) = \left(5\right) \sin \left(\frac{\pi}{6}\right) = 2.5$

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:
$0 = 2.5 + \left(- 9.8\right) \Delta t \implies \Delta t = \frac{2.5}{9.8}$

To find the distance in y we have to use the equation of distance:
$y = {y}_{0} + {v}_{0 y} t + \frac{1}{2} a {t}^{2}$
$\implies y = \left(2.5\right) \left(\frac{2.5}{9.8}\right) + \left(\frac{1}{2}\right) \left(- 9.8\right) {\left(\frac{2.5}{9.8}\right)}^{2}$
$\implies y \approx 0.9566 m$

To find the distance in x we have to use another time the equation of distance (this time there's no acceleration):
$x = {x}_{0} + {v}_{0 x} t + \frac{1}{2} a {t}^{2}$
$\implies x = \left(\frac{\sqrt{3}}{2}\right) \left(\frac{2.5}{9.8}\right)$
$\implies x \approx 0.22 m$