Chapter 1 - Chemical Tools: Experimentation and Measurement - Worked Example - Page 15: 8

The velocity of the baseball is equal to 42.5 $m/s$

** 450g = $450 \frac{1}{10^3}kg = 0.450$ $kg$ ** J = $\frac{kg*m^2}{s^2}$ Thus, we find: $E_k = \frac{1}{2}mv^2$ $406J = \frac{1}{2}(0.450kg)v^2$ $406J = (0.225kg)v^2$ $\frac{406 \frac{kg*m^2}{s^2}}{0.225kg} = v^2$ $\sqrt{\frac{406 \frac{kg*m^2}{s^2}}{0.225kg}} = v$ $v = 42.5 \frac{m}{s}$