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'Lunar Leaper' printed from https://nrich.maths.org/
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.