Question

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?

Answer 1

`t = 1.68` `"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_(0y) - g t`

to find the time.

The initial `y`-velocity `v_(0y)` is given by

`v_(0y) = v_0sinalpha_0`

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

And `g = 9.81` `"m/s"^2`

Plugging in known values, we have

`0 = color(red)(16.5)color(white)(l)color(red)("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"`