# A projectile is shot at an angle of pi/12  and a velocity of 85 m/s. How far away will the projectile land?

Dec 25, 2015

This is a standard problem about a parabolic trajectory.
Let us show that answer is ...

#### Explanation:

First of all, parabolic movement has two components: a horizontal (X axis) and a vertical (Y axis) one.

• X axis follows a rectilinear, uniform motion , described by the equation:
$x = {v}_{x} \cdot t$
where $x$ is the distance far away from the point where the projectile is launched, ${v}_{x}$ the forward speed, and $t$ the time.

• Y axis follows a rectilinear, uniformly accelerated motion, described by the equation:
$y = {v}_{0 y} \setminus \cdot t - \frac{1}{2} \cdot g \cdot {t}^{2}$
where $y$ is the height of the projectile for every moment, ${v}_{0 y}$ the initial ascent speed, $g$ the gravity acceleration $\left(g = 9 , 8 {\text{m/s}}^{2}\right)$ and $t$ the time.

The speed of projectile, $\vec{v}$, divides into two components:

$\vec{v} = \left({v}_{x} , {v}_{y}\right)$

which can be calculated:

• ${v}_{x} = v \cdot \cos \theta$
• ${v}_{y} = v \cdot \sin \theta$
where $\theta$ is the angle between trajectory and horizontal, and $v$ is projectile speed.

We are going to use only 3 of the equations above:

Step 1 of 3:
${v}_{x} = v \cdot \cos \theta = 85 \text{m/s" cdot cos (pi/12) = 82.10 "m/s}$

Step 2 of 3: let us make ${v}_{0 y} = {v}_{0} \cdot \sin \theta = 85 \text{m/s" cdot sin (pi/12) = 22.00 "m/s}$ which is the vertical speed at the beginning of the motion.
And now, we know that, at the end, the height of projectile is zero (because it lands on the floor), so $y = 0$:
$y = {v}_{0 y} \cdot t - \frac{1}{2} g {t}^{2} = 0 \rightarrow 22.00 \cdot t - 0.5 \cdot 9.8 \cdot {t}^{2} = 0 \rightarrow \left\{{t}_{1} = 0 \text{s", t_2 = 4.49 "s}\right\}$
We choose ${t}_{2}$ (${t}_{1}$ refers to the point where the projectile is shot, not where it lands.)

Step 3 of 3:
$x = {v}_{x} \cdot t = 82.10 \text{m/s" cdot 4.49 "s" = 368.63 "m}$

The projectile lands at 368.63 m from the origin.