Question

A projectile is shot from the ground at an angle of `(5 pi)/12` and a speed of `3/5 m/s`. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Answer 1

`"distance" = 0.0309` `"m"`

Explanation:

We're asked to find the distance from the launch point a particle is when it reaches its maximum height, given its initial velocity.

To do this, we can use the equations

`(v_y)^2 = (v_0sinalpha_0)^2 - 2g(Deltay)`

and

`Deltax = v_0cosalpha_0t`

to find the vertical and horizontal positions, respectively, when the particle is at its maximum height.

We realize that the instantaneous s`y`-velocity `v_y` is `0` at its maximum height, so we have

`0 = (3/5color(white)(l)"m/s")sin((5pi)/12) - 2(9.81color(white)(l)"m/s"^2)(Deltay)`

`Deltay = color(red)(0.0295` `color(red)("m"`

The time it takes for this to happen is given by

`v_y = v_0sinalpha_0 - g t`

`0 = (3/5color(white)(l)"m/s")sin((5pi)/12) - (9.81color(white)(l)"m/s"^2)t`

`t = 0.0591` `"s"`

We can now use the second equation to determine the horizontal distance `Deltax`:

`Deltax = v_0cosalpha_0t = (3/5color(white)(l)"m/s")cos((5pi)/12)(0.0591color(white)(l)"s")`

`Deltax = color(green)(0.00917` `color(green)("m"`

The distance from the launch point is thus

`r = sqrt((Deltax)^2 + (Deltay)^2) = sqrt((color(green)(0.00917color(white)(l)"m"))^2 + (color(red)(0.0295color(white)(l)"m"))^2)`

`= color(blue)(ul(0.0309color(white)(l)"m"`