# A projectile is shot from the ground at a velocity of 18 m/s at an angle of pi/6. How long will it take for the projectile to land?

Jan 5, 2016

$T = 1.8367 s$

#### Explanation:

Data:-
Velocity$= {v}_{0} = 18 \frac{m}{s}$
Angle$= \theta = \frac{\pi}{6}$
Time of flight=T=??
Acceleration due to gravity$= g = 9.8 \frac{m}{s} ^ 2$
Sol:-
We know that:
$T = \frac{2 {v}_{0} \sin \theta}{g}$
$\implies T = \frac{2 \cdot 18 S \in \left(\frac{\pi}{6}\right)}{9.8} = \frac{36 \cdot 0.5}{9.8} = \frac{18}{9.8} = 1.8367 s$
$\implies T = 1.8367 s$