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?

Result

v =  112.755 km/h

Solution:

$m = 60 \ kg \ \\ h = 50 \ m \ \\ g = 9.81 \ m/s^2 \ \\ \ \\ E = m \cdot \ g \cdot \ h = 60 \cdot \ 9.81 \cdot \ 50 = 29430 \ J \ \\ \ \\ E = \dfrac{ 1 }{ 2 } m v^2 \ \\ \ \\ v_{ 1 } = \sqrt{ \dfrac{ 2 \cdot \ E }{ m } } = \sqrt{ \dfrac{ 2 \cdot \ 29430 }{ 60 } } = 3 \ \sqrt{ 109 } \ m/s \doteq 31.3209 \ m/s \ \\ \ \\ v = v_{ 1 } \rightarrow km/h = v_{ 1 } \cdot \ 3.6 \ km/h = 112.755310296 \ km/h = 112.755 \ \text { km/h }$

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