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 result:

v =  112.755 km/h

Solution:

m=60 kg h=50 m g=9.81 m/s2  E=m⋅ g⋅ h=60⋅ 9.81⋅ 50=29430 J  E=12mv2  v1=2⋅ Em=2⋅ 2943060≐3 109 m/s≐31.3209 m/s  v=v1→km/h=v1⋅ 3.6 km/h=31.3209195267⋅ 3.6 km/h=112.755 km/h=112.755 km/hm=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=\dfrac{ 1 }{ 2 } m v^2 \ \\ \ \\ v_{1}=\sqrt{ \dfrac{ 2 \cdot \ E }{ m } }=\sqrt{ \dfrac{ 2 \cdot \ 29430 }{ 60 } } \doteq 3 \ \sqrt{ 109 } \ \text{m/s} \doteq 31.3209 \ \text{m/s} \ \\ \ \\ v=v_{1} \rightarrow km/h=v_{1} \cdot \ 3.6 \ km/h=31.3209195267 \cdot \ 3.6 \ km/h=112.755 \ km/h=112.755 \ \text{km/h}

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