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I have the question "An egg falls from a nest 3.70m above the ground. Calculate its velocity when it hits the ground and the time it takes to fall".

I know that velocity V = acceleration (a) X time (t)

I also know that the acceleration in this case is gravity so 9.81 ms$^-2$.

However, there is no time given and so I do not know how to calculate the velocity of the egg once it has hit the ground.

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    $\begingroup$ Work out the distance travelled in free fall from an initial velocity of zero, and solve for the time it takes to go $3.70$m. $\endgroup$ – hardmath Jan 19 '17 at 17:05
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    $\begingroup$ Use conservation of energy. $\endgroup$ – amd Jan 19 '17 at 20:04
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Notice that the potential energy is given by:

$$\text{E}_\text{pot}=\text{m}\cdot\text{g}\cdot\text{h}\tag1$$

And the kinetic energy:

$$\text{E}_\text{kin}=\frac{\text{m}\cdot\text{v}^2}{2}\tag2$$

So, set those equal (conservation of energy):

$$\text{E}_\text{pot}=\text{E}_\text{kin}=\text{m}\cdot\text{g}\cdot\text{h}=\frac{\text{m}\cdot\text{v}^2}{2}\space\Longleftrightarrow\space\text{v}=\pm\sqrt{2\cdot\text{g}\cdot\text{h}}\tag3$$

So, for $\text{v}$ we will get:

$$\text{v}\approx\sqrt{2\cdot9.81\cdot3.70}=\sqrt{\frac{36297}{500}}\approx8.520211265\tag4$$

And the time it takes:

$$t\approx\frac{\sqrt{\frac{36297}{500}}}{9.81}=\sqrt{\frac{740}{981}}\approx0.868523065\tag5$$

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You can determine the time that it takes for the object to hit the ground.

How? Recall that you can determine the distance traveled after $t$ seconds of fall.

This is done by the formula $d(t)=\frac{9.81t^2}{2}$ (the reason is that you want $\int \limits_{0}^t 9.81xdx$ )

So we have to solve $\frac{9.81t^2}{2}=3.70$.

After finding the value of $t$ the speed is just $9.81t$

Note: I skipped the symbols for the measurements purposefully.

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Since acceleration and distance are given, you can use $v^2 = 2as$.

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In this case, you can use one of Newton's Laws of Constant acceleration:

$$v^2=u^2+2as$$

You are trying to find the final velocity $v$. $u=0 \text{ ms}^{-1}$ is the initial velocity, $a=-g \text{ ms}^{-2}$ and $s=3.70 \text{ m}$.

Then, you can find the time taken using $v=u+at$, by using the velocity you found on the first part.

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