Substituting the given values, we have $v_{2} = \frac{2m_{1}}{m_{1}+2m_{1}}\sqrt{2gh}$.
Simplifying, we get $v_{2} = \frac{2}{3}\sqrt{2gh}$.
Substituting the values for $g$ and $h$, we get $v_{2} = \frac{2}{3}\sqrt{2(9.81 m/s^{2})(2.5 m)} = 4.73 m/s$.
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