To solve this problem, we need to use both energy conservation and potential kinetic equations.
When the energy accumulated from a certain height is released, it becomes 'motion' energy or kinetic energy. Mathematically this can be expressed as
PE = KE
[tex]mg\Delta h = \frac{1}{2}mv^2[/tex]
Where
m = mass
g = Gravitational acceleration
v = Velocity
[tex]\Delta h =[/tex] Change of the height
We know that the body, based on a reference system where the floor is the zero coordinate, starts from being 25 meters high to fall to 3 meters high, so the total difference in height would be
[tex]\Delta h = 25-3[/tex]
We also have to
[tex]g = 9.8m/s^2[/tex]
Using the previous equation we have to:
[tex]mg\Delta h = \frac{1}{2}mv^2[/tex]
[tex]g\Delta h = \frac{1}{2}v^2[/tex]
[tex]v = \sqrt{2g\Delta h}[/tex]
Replacing
[tex]v = \sqrt{2(9.8)(25-3)}[/tex]
[tex]v = 20.76m/s\approx 20.8m/s[/tex]
The correct answer is 20.8m/s
