A rock is dropped into a well 49m deep. Determine the amount of time it takes for the rock to hit the bottom of the well.

$$\\\ddot y =9.8\;\;m/s^2\\
\dot y = 9.8t\;\; m/s\\
y=\frac{9.8t^2}{2}=4.9t^2\;\;m\\
\mbox{When y=49 find t}\\
49=4.9t^2\\
10=t^2\\
t=\sqrt{10}=3.16\;\;sec\;\; \quad (to \;2\;dec\;places)$$

We have, assuming the top of the well is at a height of "0,"

-49 = -4.9t²      divide both sides by -4.9

10 = t²              take the square root of both sides

t = √10 = about 3.16 sec