# A projectile is shot from the ground at a velocity of 9 m/s and at an angle of (5pi)/6. How long will it take for the projectile to land?

Jan 6, 2016

$T = 0.9184 s$

#### Explanation:

Data:-
Muzzle velocity$=$Initial velocity$= {v}_{0} = 9 \frac{m}{s}$
Angle$= \theta = \frac{5 \pi}{6}$
Time of flight=T=??
Sol:-
$T = \frac{2 {v}_{0} S \int h \eta}{g}$

$\implies T = \frac{2 \cdot 9 \sin \left(\frac{5 \pi}{6}\right)}{9.8} = \frac{18 \cdot 0.5}{9.8} = \frac{9}{9.8} = 0.9184 s$

$\implies T = 0.9184 s$