# A projectile is shot from the ground at a velocity of 8 m/s and at an angle of (7pi)/12. How long will it take for the projectile to land?

Jul 21, 2017

$t = 1.58$ $\text{s}$

#### Explanation:

We're asked to find the time $t$ of an object's flight, given its initial velocity.

To do this, we recognize that when the object lands (assuming it launches and lands at the same height), its height $\Delta y$ will be $0$; we can use the equation

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

to find this time. For this equation,

• $\Delta y$ is the change in height ($0$)

• ${v}_{0}$ is the initial speed ($8$ $\text{m/s}$)

• ${\alpha}_{0}$ is the launch angle ($\frac{7 \pi}{12}$)

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

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

Plugging in known quantities:

$0 = \left(8 \textcolor{w h i t e}{l} {\text{m/s")sin((7pi)/12)t - 1/2(9.81color(white)(l)"m/s}}^{2}\right) {t}^{2}$

$\frac{1}{2} \left(9.81 \textcolor{w h i t e}{l} \text{m/s"^2)t^2 = (7.73color(white)(l)"m/s}\right) t$

(4.905color(white)(l)"m/s"^2)t = 7.73color(white)(l)"m/s"

t = (7.73color(white)(l)"m/s")/(4.905color(white)(l)"m/s"^2) = color(red)(1.58 color(red)("s"