# I am on the moon where gravity is 1.6 m/s/s downward. If I dropped a brick from a height of 3 m, how long did it take to hit the surface?

Aug 4, 2017

$t = 1.94$ $\text{s}$

#### Explanation:

We're asked to find the time it takes an object to hit the ground, given its height and downward acceleration.

To do this, we can use the equation

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

where

• $y$ is the final height ($0$, ground level)

• ${y}_{0}$ is the initial height (given as $3$ $\text{m}$)

• ${v}_{0 y}$ is the initial $y$-velocity ($0$ since it was dropped from a state of rest)

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

• $g$ is the acceleration due to gravity (given as $1.6$ ${\text{m/s}}^{2}$)

Plugging in known values, we have

0 = 3color(white)(l)"m" + (0)t - 1/2(1.6color(white)(l)"m/s"^2)t^2

(0.8color(white)(l)"m/s"^2)t^2 = 3color(white)(l)"m"

$t = \sqrt{\left(3 \textcolor{w h i t e}{l} \text{m")/(0.8color(white)(l)"m/s"^2)) = color(red)(ulbar(|stackrel(" ")(" "1.94color(white)(l)"s"" }\right) |}$