# If a projectile is shot at an angle of (5pi)/12 and at a velocity of 1 1 m/s, when will it reach its maximum height?

May 1, 2018

When a projectile reaches the maxima of its parabolic trajectory, ${\nu}_{y} = 0$.

Given,

${\nu}_{y} = 0$

nu_(y0) = nusintheta approx (10.6"m")/"s"

$a = g$

Now, recall,

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

Hence, it will take,

$\implies t = \frac{\Delta {\nu}_{y}}{a} \approx 1.08 \text{s}$

for the projectile to reach its maximum height.