# A projectile is shot from the ground at a velocity of 4 m/s and at an angle of (5pi)/6. How long will it take for the projectile to land?

Jul 19, 2017

$t = 0.408$ $\text{s}$

#### Explanation:

We're asked to find the time $t$ it takes a projectile to land, given its initial velocity.

When the particle lands, its height will be $0$ (assuming it launches and lands at the same height), and we can use the kinematics equation

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

where

• $\Delta y$ is the height ($0$)

• ${v}_{0 y}$ is the initial $y$-velocity, equal to

${v}_{0 y} = {v}_{0} \sin {\alpha}_{0} = \left(4 \textcolor{w h i t e}{l} \text{m/s}\right) \sin \left(\frac{5 \pi}{6}\right) = 2$ $\text{m/s}$

• $t$ is the time (what we're trying to find)

• $g$ is the acceleration due to gravity near earth's surface, $9.81$ ${\text{m/s}}^{2}$

Plugging in known values, we have

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

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

1/2(9.81color(white)(l)"m/s"^2)t = 2color(white)(l)"m/s"

t = (2color(white)(l)"m/s")/(1/2(9.81color(white)(l)"m/s"^2)) = color(red)(0.408 color(red)("s"