# What is the kinetic energy of a 0.135 kg baseball thrown at 40.0 m/s?

Use the equation, ${E}_{k} = \frac{m {v}^{2}}{2}$.
${E}_{k} = \frac{m {v}^{2}}{2}$
${E}_{k} = \frac{\left(0.135\right) {\left(40.0\right)}^{2}}{2}$.
${E}_{k} = 108 J$.