# If a projectile is shot at an angle of pi/4 and at a velocity of 16 m/s, when will it reach its maximum height??

Jan 17, 2016

1.15s

#### Explanation:

A projectile reaches its maximum height when vertical velocity is zero.

Initial velocity, u = $16 \cos \left(\frac{\pi}{4}\right) = 16 \cdot \frac{\sqrt{2}}{2} = 8 \sqrt{2}$

Taking acceleration as $a = - 9.8 m {s}^{-} 1$ then:

$v = u + a t$
$0 = 8 \sqrt{2} - 9.8 t$
$t = \frac{8 \sqrt{2}}{9.8} = 1.15 s$ (3sf)