# If a projectile is shot at a velocity of 23 m/s and an angle of pi/12, how far will the projectile travel before landing?

Jun 11, 2017

$27$ $\text{m}$

#### Explanation:

We're asked to find the horizontal range of the launched projectile with a known initial velocity.

To find this distance, we first must find the time $t$ when the object lands, using the equation

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

where

• $\Delta y$ is the change in height of the projectile, in $\text{m}$,

• ${v}_{0 y}$ is the initial $y$-velocity, in $\text{m"/"s}$. To find this, we must use the equation

${v}_{0 y} = {v}_{0} \sin \alpha$

where

• ${v}_{0}$ is the initial speed ($23 \text{m"/"s}$), and

• $\alpha$ is the initial launch angle ($\frac{\pi}{12}$)

Therefore, the initial $y$-velocity is

v_(0y) = (23"m"/"s")sin(pi/12) = 5.95"m"/"s"

• $t$ is the time, in $\text{s}$, and

• $g$ is the acceleration due to gravity near Earth's surface, 9.8"m"/("s"^2)

Plugging in $0$ for $\Delta y$ (because we must find the time when it lands at the same height as it is launched), we have

0 = (5.95"m"/"s")t - 1/2(9.8"m"/("s"^2))t^2

1/2(9.8"m"/("s"^2))t = 5.95"m"/"s"

t = color(red)(1.21 color(red)("s"

Now, to find how far horizontally it traveled, we use the equation

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

We must find the initial $x$-velocity now, using the equation

${v}_{0 x} = {v}_{0} \cos \alpha = 23 \text{m"/"s"cos(pi/12) = 22.2"m"/"s}$

The horizontal range is thus

Deltax = (22.2"m"/cancel("s"))(color(red)(1.21)cancel(color(red)("s"))) = color(blue)(27 color(blue)("m"

rounded to $2$ significant figures, the amount given in the problem.