# A projectile is shot at an angle of pi/8  and a velocity of  3 m/s. How far away will the projectile land?

May 24, 2017

$0.651 \text{m}$

#### Explanation:

We need to find the maximum range of the projectile, assuming its launch and landing heights are equal.

In terms of physics, we need to find the distance $\Delta x$ ($x$ the horizontal component) of the projectile when the change in vertical position $\Delta y$ is $0$.

The initial $y$-velocity ${v}_{0 y}$ of the projectile is

${v}_{0 y} = {v}_{0} \sin \alpha = 3 \text{m"/"s"sin(pi/8) = 1.15 "m"/"s}$

and, since we'll need this later, the initial $x$-velocity ${v}_{0 x}$ is

${v}_{0 x} = {v}_{0} \cos \alpha = 3 \text{m"/"s"cos(pi/8) = 2.77 "m"/"s}$

We can use the equation

$\Delta y = {v}_{0 y} t - \frac{1}{2} g {t}^{2}$

to find the time $t$ when the change in height $\Delta y$ is zero.

$0 = {v}_{0 y} t - \frac{1}{2} g {t}^{2}$

$\frac{1}{2} g {t}^{2} = {v}_{0 y} t$

We can see that one of these times is $t = 0$, because at this time, the motion hasn't started yet, so intuitively it hasn't changed it's height yet.

(There will always be two calculated times where the height $\Delta y$ (or $h$) is equal to $0$ (or any value for that matter), because we're in essence using the above equation which can be solved using the quadratic equation, which always yields two answers. Picture it like this: if you throw a ball upward and its maximum height is $15 \text{m}$ above the ground, there will be two times when its height is equal to, say, $10 \text{m}$: one time on the way up, and the other time on the way down.)

We can divide both sides by the variable $t$ to yield

$\frac{1}{2} g t = {v}_{0 y}$

So,

$t = \frac{2 {v}_{0 y}}{g}$

Plugging in known variables, we have

t = (2(1.15"m"/"s"))/(9.80"m"/"s"^2) = 0.235 "s"

Lastly, to find the horizontal range $\Delta x$ (or sometimes $R$), we'll use the equation

$\Delta x = {v}_{0 x} t$

Plugging in known variables yields

Deltax = (2.77"m"/"s")(0.235"s") = color(red)(0.651"m"

If you want to round this to one significant figure, the answer would technically be $0.7 \text{m}$. Ideally, measurements like those in this problem should be given with more than one significant figure to limit uncertainty.