Positional energy

What velocity in km/h must a body weighing 60 kg have for its kinetic energy to be the same as its positional energy at the height of 50 m?

Correct answer:

v =  112.7553 km/h

Step-by-step explanation:

$$\begin{gathered} m=60\text{kg} \\ h=50\text{m} \\ g=9.81{\text{m/s}}^2 \\ E=m\cdot g\cdot h=60\cdot 9.81\cdot 50=29430\text{J} \\ E=\frac{1}{2}m{v}^2 \\ {v}_1=\sqrt{\frac{2\cdot E}{m}}=\sqrt{\frac{2\cdot 29430}{60}}=3\sqrt{109}\text{m/s}\doteq 31.3209\text{m/s} \\ v=v_1\to \text{km/h}=v_1\cdot 3.6\text{km/h}=31.3209\cdot 3.6\text{km/h}=112.755\text{km/h}=112.7553\text{km/h} \end{gathered}$$

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