Question

A projectile is shot from the ground at an angle of `pi/8` and a speed of `5 /9 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" = 0.0114` `"m"` `= 1.14` `"cm"`

Explanation:

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

Vertical:

When the particle is at its maximum height, its instantaneous `y`-velocity is `0`, and we can use the equation

`(v_y)^2 = (v_0sinalpha_0)^2 - 2gh`

to find the height `h`

Plugging in known values (`v_0 = 5/9` `"m/s"`, `alpha_0 = pi/8`), we have

`0 = ((5/9color(white)(l)"m/s")sin(pi/8))^2 - 2(9.81color(white)(l)"m/s"^2)h`

`h = ul(0.00230color(white)(l)"m"`

The time `t` when this occurs is given by

`v_y = v_0sinalpha_0 - g t`

`0 = (5/9color(white)(l)"m/s")sin(pi/8) - (9.81color(white)(l)"m/s"^2)t`

`t = 0.0217` `"s"`

Horizontal:

We can find the horizontal distance covered by the equation

`x = v_0cosalpha_0t`

We found that `t = 0.0217` `"s"`, so we have

`x = (5/9color(white)(l)"m/s")cos(pi/8)(0.0217color(white)(l)"s")`

`= ul(0.0111color(white)(l)"m"`

Distance:

The distance found via

`r = sqrt(x^2 + h^2) = sqrt((0.0111color(white)(l)"m")^2 + (0.00230color(white)(l)"m")^2)`

`= color(red)(ul(0.0114color(white)(l)"m"` `= color(red)(ul(1.14color(white)(l)"cm"`