Question

A projectile is shot at a velocity of `2 m/s` and an angle of `pi/12`. What is the projectile's maximum height?

Answer 1

`"maximum height" = 0.0137` `"m"`

Explanation:

We're asked to find the maximum height reached by a projectile given its initial speed and launch angle.

At its maximum height, the object's instantaneous `y`-velocity is `0`. We can use the equation

`(v_y)^2 = (v_(0y))^2 + 2a_yDeltay`

where

• `v_y` is the `y`-velocity at a certain height `Deltay`. This value is `0` because we're trying to find its maximum height.

• `v_(0y)` is the initial `y`-velocity.

We find this using the equation

`v_(0y) = v_0sinalpha_0`

where

• `v_0` is the initial speed (given as `2` `"m/s"`)

• `alpha_0` is the launch angle, given as `pi/12`

Therefore,

`v_(0y) = (2color(white)(l)"m/s")(sin(pi/12)) = color(red)(0.518` `color(red)("m/s"`

• `a_y` for any object in free-fall is `-g`, equal to `-9.81` `"m/s"^2`

• `Deltay` is the height (relative to original height, which is taken to be `0`), which is what we're trying to find

A list of variables we're using:

`v_y = 0`

`v_(0y) = color(red)(0.518` `color(red)("m/s"`

`a_y = -9.81` `"m/s"^2`

`Deltay = ?`

Plugging in known values into the equation, we have

`0 = (color(red)(0.518color(white)(l)"m/s"))^2 - 2(9.81color(white)(l)"m/s"^2)(Deltay)`

`(19.62color(white)(l)"m/s"^2)(Deltay) = 0.268` `"m"^2"/s"^2`

`Deltay = (0.268color(white)(l)"m"^2"/s"^2)/(19.62color(white)(l)"m/s"^2) = color(blue)(0.0137` `color(blue)("m"`

The maximum height reached by this projectile is thus `color(blue)(0.0137` `sfcolor(blue)("meters"`, which makes some sense because it was launched (a) with a small initial speed and (b) with a low angle relative to the horizontal (equivalent to `15^"o"`).