# If a projectile is shot at an angle of (7pi)/12 and at a velocity of 21 m/s, when will it reach its maximum height??

Apr 30, 2018

At the maxima of a projectile's parabolic trajectory, ${\nu}_{y} = 0$.

Given,

nu_(y0) = nusin(theta) = (20.2"m")/"s"
${\nu}_{\text{y}} = 0$
${a}_{y} = g$

Recall,

$\nu = {\nu}_{0} + a t$

Hence, it will take,

$\implies t = \frac{\Delta \nu}{a} \approx 2.07 \text{s}$

for the projectile to reach its maximum height.