# Question 2e587

Aug 7, 2017

$t = 7.14$ $\text{s}$

#### Explanation:

We're asked to find the time, in seconds, it takes an object to fall to Earth's surface from a height of $250$ $\text{m}$.

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 $250$ $\text{m}$)

• ${v}_{0 y}$ 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)

• $g = 9.81$ ${\text{m/s}}^{2}$

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

$y = {y}_{0} - \frac{1}{2} g {t}^{2}$

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

$y - {y}_{0} = - \frac{1}{2} g {t}^{2}$

$- 2 \left(y - {y}_{0}\right) = g {t}^{2}$

${t}^{2} = \frac{- 2 \left(y - {y}_{0}\right)}{g}$

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)#