We are asked to determine the final velocity of an object that is dropped from a window. To do that we will use the following equation:
[tex]v^2_f-v^2_0=2gh[/tex]where "vf" is the final velocity, "v0" is the initial velocity, "g" is the acceleration of gravity, and "h" is the height. Since the object is dropped the initial velocity is zero. Replacing we get:
[tex]v^2_f=2(9.8\frac{m}{s^2})(5m)[/tex]Solving the operations:
[tex]v^2_f=98\frac{m^2}{s^2}[/tex]Taking square root on both sides:
[tex]\begin{gathered} v_f=\sqrt[]{98\frac{m^2}{s^2}} \\ v_f=9.9\text{ m/s} \end{gathered}[/tex]Therefore, the final velocity is 9.9 meters per second.
