Question

Question #2e587

Answer 1

`t = 7.14` `"s"`

Explanation:

We're asked to find the time, in seconds, it takes an object to fall to Earth's surface from a height of `250` `"m"`.

To do this, we can use the kinematics equation

`ul(y = y_0 + v_(0y)t - 1/2g t^2`

where

• `y` is the height at time `t` (which is `0`, ground-level)

• `y_0` is the initial height (given as `250` `"m"`)

• `v_(0y)` is the initial velocity (it dropped from a state of rest, so this is `0`)

• `t` is the time (what we're trying to find)

• `g = 9.81` `"m/s"^2`

Sine the initial `y`-velocity is `0`, we can change the equation to

`y = y_0 - 1/2g t^2`

Let's solve this for our unknown variable, `t`:

`y-y_0= -1/2g t^2`

`-2(y-y_0) = g t^2`

`t^2 = (-2(y-y_0))/g`

`color(red)(t = sqrt((-2(y-y_0))/g)`

Plugging in known values:

`t = sqrt((-2(0-250color(white)(l)"m"))/(9.81color(white)(l)"m/s"^2)) = color(blue)(ulbar(|stackrel(" ")(" "7.14color(white)(l)"s"" ")|)`

So ultimately, if you're ever given a situation where you're asked to find the time it takes an object to fall a certain distance (with `0` initial velocity), just use the simplified equation

`color(red)(t = sqrt((2*"height")/g)`