# What is the terminal velocity of an object which has been dropped after it has fallen 10 meters?

Jun 11, 2015

$10 \sqrt{2}$ meters/second

#### Explanation:

Acceleration due to gravity = 9.8 "meters"/("second")^2
However, because I'm lazy, I will assume that it is close enough to 10 "meters"/("second")^2 to not make any significant difference.

Assuming an initial velocity of $0$

Distance traveled can be computed as
$\textcolor{w h i t e}{\text{XXXX}}$$d = \frac{1}{2} a {t}^{2}$ where $a$ is the accelration and $t$ is the time.

We are told the distance is 10 meters and using a = 10 "m"/(s"^2)

$\textcolor{w h i t e}{\text{XXXX}}$$10 m = \frac{1}{2} \cdot 10 \frac{m}{\sec} ^ 2 \cdot {t}^{2} \sec$
$\textcolor{w h i t e}{\text{XXXX}}$$\Rightarrow {t}^{2} = 2 \text{ sec}$
$\textcolor{w h i t e}{\text{XXXX}}$$\Rightarrow t = \sqrt{2} \sec .$

With an initial velocity of $0$ and an acceleration of $10 \frac{m}{{s}^{2}}$
the terminal velocity after $\sqrt{2}$ sec. will be $10 \sqrt{2}$ m/sec