# A projectile is shot from the ground at a velocity of 56 m/s and at an angle of (5pi)/6. How long will it take for the projectile to land?

Feb 2, 2016

$\approx 5.71$ sec

#### Explanation:

Find the vertical and horizontal components of the velocity. Gravity only acts on the vertical component, so the time it takes for the projectile to rise and fall again is the time it takes for the projectile to land.
Let the initial vertical component be ${v}_{v i}$ and the angle of projection $\theta$
Then ${v}_{v i} = v \sin \theta$
The final vertical velocity ${v}_{v f}$ is zero

${v}_{v f} = {v}_{v i} - g \cdot t$ where $g$ is the acceleration due to gravity and $t$ is the time taken to reach the top of the trajectory.
$0 = 56 \sin \left(\frac{5 \pi}{6}\right) - 9.8 \cdot t$
$t = \frac{56 \sin \left(\frac{5 \pi}{6}\right)}{9.8}$

The time taken for the projectile to land is $2 t$
$2 t = \frac{2 \cdot 56 \cdot \sin \left(\frac{5 \pi}{6}\right)}{9.8} \approx 5.71$ sec