Question

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?

Answer 1

`t = 0.0883` `"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_0sinalpha_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` `"m/s"`)

• `alpha_0` is the launch angle (given as `(2pi)/3`)

• `g` is the acceleration due to gravity near earth's surface (`9.81` `"m/s"^2`)

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

Plugging in known values, we have

`0 = (1color(white)(l)"m/s")sin((2pi)/3) - (9.81color(white)(l)"m/s"^2)t`

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