# If a projectile is shot at an angle of (2pi)/3 and at a velocity of 19 m/s, when will it reach its maximum height?

Jul 1, 2017

$t = 1.68$ $\text{s}$

#### Explanation:

We're asked to find the time when a projectile reaches its maximum height, given its initial velocity.

When the projectile is at its maximum height, its instantaneous $y$-velocity is zero.

We can use the equation

${v}_{y} = {v}_{0 y} - g t$

to find the time.

The initial $y$-velocity ${v}_{0 y}$ is given by

${v}_{0 y} = {v}_{0} \sin {\alpha}_{0}$

v_(0y) = (19color(white)(l)"m/s")sin((2pi)/3) = color(red)(16.5 color(red)("m/s"

And $g = 9.81$ ${\text{m/s}}^{2}$

Plugging in known values, we have

$0 = \textcolor{red}{16.5} \textcolor{w h i t e}{l} \textcolor{red}{{\text{m/s") - (9.81color(white)(l)"m/s}}^{2}} t$

t = (color(red)(16.5)color(white)(l)color(red)("m/s"))/(9.81color(white)(l)"m/s"^2) = color(blue)(1.68 color(blue)("s"