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

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

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Answer 1 · 2015-06-11T15:27:00

$10 \sqrt{2}$ $10sqrt(2)$ meters/second

Explanation:

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

Assuming an initial velocity of $0$ $0$

Distance traveled can be computed as
$XXXX$ $color(white)("XXXX")$ $d = \frac{1}{2} a t^{2}$ $d = 1/2at^2$ where $a$ $a$ is the accelration and $t$ $t$ is the time.

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

$XXXX$ $color(white)("XXXX")$ $10 m = \frac{1}{2} \cdot 10 \frac{m}{{sec}^{2} \cdot} t^{2} sec$ $10 m = 1/2 * 10 m/sec^2* t^2 sec$
$XXXX$ $color(white)("XXXX")$ $\Rightarrow t^{2} = 2 sec$ $rArr t^2 = 2 " sec"$
$XXXX$ $color(white)("XXXX")$ $\Rightarrow t = \sqrt{2} sec .$ $rArr t =sqrt(2) sec.$

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

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

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