# If a projectile is shot at a velocity of 11 m/s and an angle of pi/4, how far will the projectile travel before landing?

Jul 11, 2017

$\Delta x = 12.3$ $\text{m}$

#### Explanation:

We're asked to find the horizontal range of a projectile given its initial speed and launch angle.

To do this, we can first find the time $t$ when it has a height of $0$, using the equation

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

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

${v}_{0 y} = {v}_{0} \sin {\alpha}_{0} = \left(11 \textcolor{w h i t e}{l} \text{m/s}\right) \sin \left(\frac{\pi}{4}\right) = 7.78$ $\text{m/s}$

The change in height $\Delta y$ is $0$, because we're trying to find the time $t$ at this height. Plugging in known values, we have

$0 = \left(7.78 \textcolor{w h i t e}{l} {\text{m/s")t - 1/2(9.81color(white)(l)"m/s}}^{2}\right) {t}^{2}$

$\left(4.905 \textcolor{w h i t e}{l} \text{m/s"^2)t^2 = (7.78color(white)(l)"m/s}\right) t$

(4.905color(white)(l)"m/s"^2)t = 7.78color(white)(l)"m/s"

t = color(red)(1.59 color(red)("s"

We can now use the equation

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

to find the horizontal range, $\Delta x$

The initial $x$-velocity ${v}_{0 x}$ is

${v}_{0 x} = {v}_{0} \cos {\alpha}_{0} = \left(11 \textcolor{w h i t e}{l} \text{m/s}\right) \cos \left(\frac{\pi}{4}\right) = 7.78$ $\text{m/s}$

We then have:

Deltax = (7.78"m"/(cancel("s")))(1.59cancel("s")) = color(blue)(12.3 color(blue)("m"