# If a projectile is shot at an angle of (pi)/8 and at a velocity of 21 m/s, when will it reach its maximum height?

Apr 1, 2018

The time is $= 0.82 s$

#### Explanation:

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

The initial velocity is ${u}_{0} = \left(21\right) \sin \left(\frac{1}{8} \pi\right) m {s}^{-} 1$

At the greatest height, $v = 0 m {s}^{-} 1$

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

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

Applying the equation of motion

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

The time is $t = \frac{v - u}{- g} = \frac{0 - \left(21\right) \sin \left(\frac{1}{8} \pi\right)}{- 9.8} = 0.82 s$