# A projectile is shot at an angle of pi/4  and a velocity of  39 m/s. How far away will the projectile land?

Aug 10, 2017

$\Delta x \approx 155 \text{m}$

#### Explanation:

We can use kinematics and vector decomposition to calculate the range of the projectile.

We are given the following information:

• $\mapsto {v}_{i} = 39 \text{ m"//"s}$

• $\mapsto \theta = \frac{\pi}{4}$

To calculate the horizontal distance the projectile follows, we can use the following kinematic equation:

$\Delta x = {v}_{i x} \Delta t + \frac{1}{2} {a}_{x} \Delta {t}^{2}$

We assume gravity is the only force acting on the projectile (e.g. no air resistance). Because the projectile is only under the influence of gravity, it has a constant vertical acceleration of $- 9.8 {\text{m"//"s}}^{2} ,$ and there is no acceleration horizontally (we have no horizontal forces). This reduces our equation to the familiar equation for a constant velocity:

$\textcolor{p u r p \le}{\Delta x = {v}_{i x} \Delta t}$

Which we are likely used to seeing as $v = \frac{\Delta s}{\Delta t}$

We will need both the parallel (x) component of the velocity and the the total flight time of the projectile. We will have to use the vertical components to find the flight time.

As mentioned above, the vertical acceleration is ${a}_{y} = 9.8 {\text{m"//"s}}^{2}$. We also know that at the very top of the object's motion the velocity will be instantaneously zero as it changes direction, so for the first half of the flight time, ${v}_{f y} = 0$.

A convenient fact is that, provided the object lands at the same altitude it was launched from, $\textcolor{b l u e}{{t}_{\text{rise")=t_("fall}}}$. Therefore, we can calculate the flight time of the projectile for the first half of the motion and multiply it by two to find the total flight time.

We can use the following kinematic equation to find $\Delta t$:

$\cancel{{v}_{y f}} = {v}_{y i} + {a}_{y} \Delta t$

$\implies \Delta t = \frac{- {v}_{y i}}{a}$

We will now find the parallel (x) and perpendicular (y) components of the velocity.

Here is a diagram for the first (initial) velocity vector, ${v}_{i}$: Using trigonometry, we see that:

$\sin \left(\theta\right) = \text{opposite"/"hypotenuse}$

$\implies \sin \left(\theta\right) = {v}_{y} / v$

$\implies \textcolor{c r i m s o n}{{v}_{y} = v \sin \left(\theta\right)}$

Similarly, we find that $\textcolor{c r i m s o n}{{v}_{x} = v \cos \left(\theta\right)}$

So, we have:

${v}_{y} = 39 \sin \left(\frac{\pi}{4}\right)$

$= \frac{39 \sqrt{2}}{2} \text{m"//"s}$

$\textcolor{\mathrm{da} r k b l u e}{\approx 27.58 \text{ m"//"s}}$

${v}_{x} = 39 \cos \left(\frac{\pi}{4}\right)$

$= \frac{39 \sqrt{2}}{2} \text{m"//"s}$

$\textcolor{\mathrm{da} r k b l u e}{\approx 27.58 \text{ m"//"s}}$

We can now solve for $\Delta t$:

$\Delta t = \frac{- {v}_{y i}}{a}$

$= \left(- 27.58 {\text{ m"//"s")/(-9.81" m"//"s}}^{2}\right)$

$= 2.81 \text{s}$

Therefore, the total flight time is $\left(2.81 \cdot 2\right) = \textcolor{p u r p \le}{5.62 \text{s}}$

Revisiting our first kinematic for the range...

$\Delta x = {v}_{i x} \Delta t$

$= \left(27.58 \text{m"//"s")(5.62"s}\right)$

$= 155.078 \text{m}$

$\textcolor{c r i m s o n}{\approx 155 \text{m}}$