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.
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.
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.
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:
$$t\approx\frac{\sqrt{\frac{36297}{500}}}{9.81}=\sqrt{\frac{740}{981}}\approx0.868523065\tag5$$
Since acceleration and distance are given, you can use $v^2 = 2as$.