Answer:
The speed of the cat when it hits the ground is approximately 7.586 meters per second.
Explanation:
By Principle of Energy Conservation and Work-Energy Theorem, we have that initial potential gravitational energy of the cat ([tex]U_{g}[/tex]), in joules, is equal to the sum of the final translational kinetic energy ([tex]K[/tex]), in joules, and work losses due to air resistance ([tex]W_{l}[/tex]), in joules:
[tex]U_{g} = K +W_{l}[/tex] (1)
By definition of potential gravitational energy, translational kinetic energy and work, we expand the equation presented above:
[tex]m \cdot g\cdot h = \frac{1}{2}\cdot m \cdot v^{2}+W_{l}[/tex] (2)
Where:
[tex]m[/tex] - Mass of the cat, in kilograms.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]h[/tex] - Initial height of the cat, in meters.
[tex]v[/tex] - Final speed of the cat, in meters per second.
If we know that [tex]m = 2.70\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]h = 5.20\,m[/tex] and [tex]W_{l} = 120\,J[/tex], then the final speed of the cat is:
[tex]v = \sqrt{\frac{2\cdot (m\cdot g\cdot h-W_{l})}{m} }[/tex]
[tex]v = \sqrt{2\cdot g\cdot h-\frac{W_{l}}{m} }[/tex]
[tex]v \approx 7.586\,\frac{m}{s}[/tex]
The speed of the cat when it hits the ground is approximately 7.586 meters per second.
