Answer:
The mass of the banana is m and it is at height h.
Applying the Law of Conservation of Energy
Total Energy before fall = Total Energy after fall
[tex]E_{i}[/tex] = [tex]E_{f}[/tex]
Here, total energy is the sum of kinetic energy and potential energy
[tex]K.E_{i}[/tex] + [tex]P.E_{i}[/tex] = [tex]K.E_{f}[/tex] + [tex]P.E_{f}[/tex] (a)
When banana is at height h, it has
[tex]K.E_{i}[/tex] = 0 and [tex]P.E_{i}[/tex] = mgh
and when it reaches the river, it has
[tex]K.E_{f}[/tex] = 1/2m[tex]v^{2} [/tex] and [tex]P.E_{f}[/tex] = 0
Putting the values in equation (a)
0 + mgh = 1/2m[tex]v^{2} [/tex] + 0
mgh = 1/2m[tex]v^{2} [/tex]
cutting 'm' from both sides
gh = 1/2[tex]v^{2} [/tex]
v = [tex]\sqrt{2gh} [/tex]
Hence, the velocity of banana before hitting the water is
v = [tex]\sqrt{2gh} [/tex]
