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Sagot :
When a ball is at a particular height it possess potential energy which is given as,
[tex]U=\text{mgh}[/tex]The potential energy is converted into kinetic energy of the ball. According to conservation of energy,
[tex]K=U[/tex]which can be further given as,
[tex]K=\text{mgh}[/tex]Substitute the known values,
[tex]\begin{gathered} K=(45.93\text{ g)(}\frac{1\text{ kg}}{1000\text{ g}})(9.8m/s^2)(31\text{ cm)(}\frac{1\text{ m}}{100\text{ cm}})(\frac{1\text{ J}}{1kgm^2s^{-2}^{}}) \\ =0.140\text{ J} \end{gathered}[/tex]Thus, the kinetic energy of the ball is 0.140 J.
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