# A projectile is shot from the ground at a velocity of 2 m/s at an angle of (5pi)/12. How long will it take for the projectile to land?

Mar 27, 2017

The time to land $= 0.4 s$

#### Explanation:

Solving in the vertical direction ${\uparrow}^{+}$

$u = {u}_{0} \sin \theta = 2 \sin \left(\frac{5}{12} \pi\right) m {s}^{-} 1$

$a = - g = - 9.8 m {s}^{-} 2$

$v = 0$ at the maximum height

We apply the equation

$v = u + a t$

$0 = 2 \sin \left(\frac{5}{12} \pi\right) - g \cdot t$

$t = \frac{2 \sin \left(\frac{5}{12} \pi\right)}{g} = 0.2 s$

This is the time to reach the maximum height

The time to land $= 0.2 \cdot 2 = 0.4 s$