# If a projectile is shot at an angle of (2pi)/3 and at a velocity of 1 m/s, when will it reach its maximum height?

Jul 21, 2017

$t = 0.0883$ $\text{s}$

#### Explanation:

We're asked to find the time $t$ when a particle reaches its maximum height, given its initial velocity.

To do this, we can use the equation

${v}_{y} = {v}_{0} \sin {\alpha}_{0} - g t$

where

• ${v}_{y}$ is the velocity at time $t$ (this will be $0$, since at a particle's maximum height its instantaneous $y$-velocity is $0$)

• ${v}_{0}$ is the initial speed (give as $1$ $\text{m/s}$)

• ${\alpha}_{0}$ is the launch angle (given as $\frac{2 \pi}{3}$)

• $g$ is the acceleration due to gravity near earth's surface ($9.81$ ${\text{m/s}}^{2}$)

• $t$ is the time (what we're trying to find)

Plugging in known values, we have

$0 = \left(1 \textcolor{w h i t e}{l} {\text{m/s")sin((2pi)/3) - (9.81color(white)(l)"m/s}}^{2}\right) t$

t = (0.866color(white)(l)"m/s")/(9.81color(white)(l)"m/s"^2) = color(red)(0.0883 color(red)("s"