Question

If a projectile is shot at a velocity of `23 m/s` and an angle of `pi/12`, how far will the projectile travel before landing?

Answer 1

`27` `"m"`

Explanation:

We're asked to find the horizontal range of the launched projectile with a known initial velocity.

To find this distance, we first must find the time `t` when the object lands, using the equation

`Deltay = v_(0y)t - 1/2g t^2`

where

• `Deltay` is the change in height of the projectile, in `"m"`,

• `v_(0y)` is the initial `y`-velocity, in `"m"/"s"`. To find this, we must use the equation

`v_(0y) = v_0sinalpha`

where

• `v_0` is the initial speed (`23"m"/"s"`), and

• `alpha` is the initial launch angle (`pi/12`)

Therefore, the initial `y`-velocity is

`v_(0y) = (23"m"/"s")sin(pi/12) = 5.95"m"/"s"`

• `t` is the time, in `"s"`, and

• `g` is the acceleration due to gravity near Earth's surface, `9.8"m"/("s"^2)`

Plugging in `0` for `Deltay` (because we must find the time when it lands at the same height as it is launched), we have

`0 = (5.95"m"/"s")t - 1/2(9.8"m"/("s"^2))t^2`

`1/2(9.8"m"/("s"^2))t = 5.95"m"/"s"`

`t = color(red)(1.21` `color(red)("s"`

Now, to find how far horizontally it traveled, we use the equation

`Deltax = v_(0x)t`

We must find the initial `x`-velocity now, using the equation

`v_(0x) = v_0cosalpha = 23"m"/"s"cos(pi/12) = 22.2"m"/"s"`

The horizontal range is thus

`Deltax = (22.2"m"/cancel("s"))(color(red)(1.21)cancel(color(red)("s"))) = color(blue)(27` `color(blue)("m"`

rounded to `2` significant figures, the amount given in the problem.