# A projectile is shot from the ground at a velocity of 54 m/s and at an angle of (7pi)/12. How long will it take for the projectile to land?

Oct 14, 2016

$t \approx 10.6 s$

#### Explanation:

Write the equation for the y coordinate as a function of time:

$y \left(t\right) = \left(54 \frac{m}{s}\right) \sin \left(\frac{7 \pi}{12}\right) t - \left(\frac{1}{2}\right) \left(9.8 \frac{m}{s} ^ 2\right) {t}^{2}$

To find the times when the projectile is on the ground we set $y \left(t\right) = 0$:

$0 = \left(54 \frac{m}{s}\right) \sin \left(\frac{7 \pi}{12}\right) t - \left(\frac{1}{2}\right) \left(9.8 \frac{m}{s} ^ 2\right) {t}^{2}$

$t \left(\left(54 \frac{m}{s}\right) \sin \left(\frac{7 \pi}{12}\right) - \left(\frac{1}{2}\right) \left(9.8 \frac{m}{s} ^ 2\right) t\right) = 0$

We know that the projectile is launched from the ground so $t = 0$ is an extraneous root and, therefore, we can divide the equation by t:

$\left(54 \frac{m}{s}\right) \sin \left(\frac{7 \pi}{12}\right) - \left(\frac{1}{2}\right) \left(9.8 \frac{m}{s} ^ 2\right) t = 0$

Solve the remaining factor for t:

$t = 2 \frac{54 \frac{m}{s}}{9.8 \frac{m}{s} ^ 2} \sin \left(\frac{7 \pi}{12}\right)$

$t \approx 10.6 s$