A projectile is shot from the ground at a velocity of 27 m/s at an angle of (5pi)/12. How long will it take for the projectile to land?

Aug 11, 2018

The time to land is $= 5.32 s$

Explanation:

Resolving in the vertical direction and applying the equation of motion

$s = u t + \frac{1}{2} a {t}^{2}$

The distance travelled is $s = 0$

The initial velocity is $u = 27 \sin \left(\frac{5}{12} \pi\right)$

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

$27 \sin \left(\frac{5}{12} \pi\right) \cdot t - \frac{1}{2} \cdot 9.8 \cdot {t}^{2} = 0$

$t = 0$ which is the initial conditions

$t = \frac{27 \sin \left(\frac{5}{12} \pi\right)}{4.9} = 5.32 s$