# A projectile is shot from the ground at an angle of (5 pi)/12  and a speed of 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?

Jan 8, 2016

$.230$ meters from the starting point.

#### Explanation:

We should start by breaking down the initial velocity into its $x$ and $y$ components. We can find the projection of $\vec{V}$ onto the $x$ and $y$ axis by using $\cos \theta$ and $\sin \theta$ respectively.

$\vec{V} = \vec{V} \cdot \hat{x} + V \cdot \hat{y}$

$= \left(V \cos \theta\right) \hat{x} + \left(V \sin \theta\right) \hat{y}$

$= 3 \cos \left(\frac{5 \pi}{12}\right) \hat{x} + 3 \sin \left(\frac{5 \pi}{12}\right) \hat{y}$

$= .776 \hat{x} + 2.898 \hat{y}$

Note that the units are $\text{m/s}$. The initial velocity is mostly in the $y$ direction, so we shouldn't expect the projectile to go very far. Now we can calculate how long it takes for the projectile to reach the highest point. At its highest, the vertical velocity of the projectile will be zero. We can use the following equation to find at what time the velocity is zero.

${V}_{y} \left(t\right) = a t + {V}_{\circ} = 0$

$t = - {V}_{\circ} / a$

The acceleration due to gravity is $- 9.8 \frac{m}{s} ^ 2$.

$t = - \frac{2.989}{- 9.8}$

$t = .296$

So the projectile takes $.296 \text{ seconds}$ to reach the top of its flight. Now we can plug this into the equation of motion for the $x$ direction. Since there is no net force on the projectile in the horizontal direction, the acceleration will be zero. Also, we are starting at the origin, so the initial $x$ position will be zero.

$x \left(t\right) = \frac{1}{2} {\cancel{a}}^{0} {t}^{2} + {V}_{x} t + {\cancel{{x}_{\circ}}}^{0}$

$x \left(.296\right) = \left(.776\right) \left(.296\right)$

$x \left(.296\right) = .230$

The projectile will move about $.230 \text{ m}$ from the starting point when it reaches the top of its flight.