Question

A projectile is shot from the ground at a velocity of `4 m/s` and at an angle of `(5pi)/6`. How long will it take for the projectile to land?

Answer 1

`0.4` `"s"`

Explanation:

We're asked to find the time `t` when the projectile lands with a known initial velocity and direction. In terms of physics, we need to find the time when the change in `y` (the height) is equal to `0`.

We can use the equation

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

to find the time `t` when `Deltay` is zero,

where

• `Deltay` is the change in `y`-position (height) of the object, in `"m"`

• `v_(0y)` is the initial `y`-velocity of the object, in `"m"/"s"`

• `t` is the time, in `"s"`, and

• `g` is the acceleration due to gravity near Earth's surface, `9.80"m"/("s"^2)`.

To find the initial `y`-velocity, we can use the equation

`v_(0y) = v_0sinalpha`

where

• `v_0` is the magnitude of the initial velocity, in `"m"/"s"`, and

• `alpha` is the initial launch angle.

Plugging in our known variables, we have

`v_(0y) = v_0sinalpha = (4"m"/"s")sin((5pi)/6) = color(red)(2"m"/"s"`

Now that we have our known variables necessary, let's plug them back into the original equation to find the times when the height is `0` (there are two times because the height is zero when it is first launched as well):

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

`0 = (color(red)(2"m"/"s"))t - 1/2(9.80"m"/("s"^2))t^2`

`1/2(9.80"m"/("s"^2))t^2 = (2"m"/"s")t`

We'll eliminate the `t = 0` option by dividing both sides by `t`:

`(4.90"m"/("s"^2))t = (2"m"/"s")`

`t = (2cancel("m")/cancel("s"))/(4.90cancel("m")/("s"^cancel(2))) = color(blue)(0.4"s"`

rounded to one significant figure, which is the amount given in the problem.

Thus, the object will land again ad `t = 0.4"s"`