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 50 m?

Correct answer:

v =  112.7553 km/h

Step-by-step explanation:

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

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