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Andaleeb of Woodhouse Sixth Form College sent in this correct solution. He takes account of the fact that pole vaulters only have to lift their center of gravity to roughly the height of the bar in order to clear it. Another correct solution was received from Christopher of Sidcup, Kent.
Since the man is 2m tall, his centre of mass is roughly 1m from the ground.
Let $h =$ height from the centre of mass to the highest point of clearing. Then on the Earth, $h = 5 - 1 = 4$ metres. Using conservation of energy, we get: $$\frac{1}{2}mv^2 = 4mg$$ where $v$ is the linear velocity moving horizontally only. Thus $$v^2 = 8g$$ Since the man is moving horizontally only (no vertical component of velocity is present) his velocity is not affected by gravity. Thus, $$v_{Earth} = v_{moon}$$ Applying conservation of energy on the moon, we get:
Lunar Leaper
Age 16 to 18
Challenge Level 
Andaleeb of Woodhouse Sixth Form College sent in this correct solution. He takes account of the fact that pole vaulters only have to lift their center of gravity to roughly the height of the bar in order to clear it. Another correct solution was received from Christopher of Sidcup, Kent.
Since the man is 2m tall, his centre of mass is roughly 1m from the ground.
Let $h =$ height from the centre of mass to the highest point of clearing. Then on the Earth, $h = 5 - 1 = 4$ metres. Using conservation of energy, we get: $$\frac{1}{2}mv^2 = 4mg$$ where $v$ is the linear velocity moving horizontally only. Thus $$v^2 = 8g$$ Since the man is moving horizontally only (no vertical component of velocity is present) his velocity is not affected by gravity. Thus, $$v_{Earth} = v_{moon}$$ Applying conservation of energy on the moon, we get:
\begin{eqnarray} \frac{1}{2}m(8g) &=&
m\frac{g}{6}h \\ \Rightarrow h &=& 24. \\
\end{eqnarray}
Thus on the moon, the pole vaulter can clear a pole of height 24 +
1 = 25m. You may also like
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