# If a projectile is shot at an angle of (5pi)/12 and at a velocity of 27 m/s, when will it reach its maximum height?

Jul 27, 2017

The time is $= 2.66 s$

#### Explanation:

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

We apply the equation of motion

$v = u + a t$

The initial velocity is $u = 27 \sin \left(\frac{5}{12} \pi\right) m {s}^{-} 1$

The final velocity (at the maximum height) is $v = 0 m {s}^{-} 1$

The acceleration is $a = - g m {s}^{-} 2$

Therefore,

the time is

$t = \frac{v - u}{g}$

$t = 0 - 27 \sin \frac{\frac{5}{12} \pi}{- 9.8} = 2.66 s$