Question

A stone is dropped freely, while another thrown vertically downward with an initial velocity of 2m/s from the same point, simultaneously. What is the time required by them to have a distance of 22 m between them?

Answer 1

`t = 11` `"s"`

Explanation:

We're asked to find the time `t` when the distance between the two stones, which we'll call `Deltay`, is `22` `"m"`.

The equation for the position of stone 1 (the one dropped from rest) is

`Deltay_1 = -1/2g t^2`

(the initial `y`-velocity `v_(0y)` is `0`)

and that of stone 2 (the one launched downward) is

`Deltay_2 = v_(0y)t - 1/2g t^2`

Plugging in the known values, these equations become

`Deltay_1 = (-4.9"m"/("s"^2))t^2`

`Deltay_2 = (-2"m"/"s")t - (4.9"m"/("s"^2))t^2`

At any given time, the position of stone 2 will always be farther than that of stone 1, because it was launched with an initial velocity downward. Since we want to find when the distance between them is `22` `"m"`, we can say

`Deltay_2 = Deltay_1 - 22` `"m"` (minus because it's farther downward)

Substituting the above equations for `Deltay_2` and `Deltay_1`, we have

`(-2"m"/"s")t - (4.9"m"/("s"^2))t^2 = (-4.9"m"/("s"^2))t^2 - 22` `"m"`

`(-2"m"/"s")t = -22` `"m"`

`t = color(red)(11` `color(red)("s"`

Thus, the distance between the objects will be `22` meters after `color(red)(11` `sfcolor(red)("seconds"`.

We can check to see if this answer is correct by finding their positions at time `t = 11` `"s"`:

`Deltay_1 = -1/2g t^2 = (-4.9"m"/("s"^2))(11"s")^2 = -593` `"m"`

`Deltay_2 = v_(0y)t - 1/2g t^2 = (-2"m"/"s")(11"s") - (4.9"m"/("s"^2))(11"s")^2`

`= -615` `"m"`

`"Distance between them" = Deltay_1 - Deltay_2 = -593"m" - (-615"m")`

`= color(blue)(22` `color(blue)("m"`