Question

A projectile is shot from the ground at an angle of `(5 pi)/12` and a speed of `8 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" = 3.45` `"m"`

Explanation:

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

To do this, we need to find the vertical and horizontal components of the position at this point. For reference during our work, the components of the initial velocity are

`v_(0x) = (8color(white)(l)"m/s")cos((5pi)/12) = 2.07` `"m/s"`

`v_(0y) = (8color(white)(l)"m/s")sin((5pi)/12) = 7.73` `"m/s"`

Vertical Position

To find the height `Deltay` when the projectile is at its maximum height, we can use the equation

`(v_y)^2 = (v_(0y))^2 + 2a_y(Deltay)`

The acceleration `a_y` is `-g`. At its maximum height, the instantaneous velocity `v_y = 0`, so we have

`0 = (7.73color(white)(l)"m/s")^2 - 2(9.81color(white)(l)"m/s"^2)(Deltay)`

`2(9.81color(white)(l)"m/s"^2)(Deltay) = 59.7` `"m"^2"/s"^2`

`Deltay = (59.7color(white)(l)"m"^2"/s"^2)/(2(9.81color(white)(l)"m/s"^2)) = color(red)(3.04` `color(red)("m"`

The time `t` when this occurs is given by

`v_y = v_(0y) - g t`

`t = (v_y - v_(0y))/(-g) = (0 - 7.73color(white)(l)"m/s")/(-9.81color(white)(l)"m/s"^2) = 0.788` `"s"`

We'll use this in calculating the horizontal component below.

Horizontal Position

The horizontal position `Deltax` is given by

`Deltax = v_(0x)t`

We found `t` above, so plugging in known values, we have

`Deltax = (2.07color(white)(l)"m/s")(0.788color(white)(l)"s") = color(green)(1.63` `color(green)("m"`

Distance

The distance `r` from the starting point is given by

`r = sqrt((Deltax)^2 + (Deltay)^2)`

Plugging in:

`r = sqrt((color(green)(1.63)color(white)(l)color(green)("m"))^2 + (color(red)(3.04color(white)(l)color(red)("m")))^2) = color(blue)(3.45` `color(blue)("m"`