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

Dec 27, 2017

The time is $= 0.106 s$

#### Explanation:

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

The initial velocity is ${u}_{0} = 2 \sin \left(\frac{1}{12} \pi\right) m {s}^{-} 1$

The time to reach the greatest height is $= t s$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

Applying the equation of motion

$v = u + a t = u - g t$

The time is $t = \frac{v - u}{- g} = \frac{0 - 2 \sin \left(\frac{1}{12} \pi\right)}{- 9.8} = 0.053 s$

The time taken to land is twice the time to reach the greatest height

$T = 2 t = 2 \cdot 0.053 = 0.106 s$