Gravity on the Moon is about 1/6th that on the Earth. A pole-vaulter 2 metres tall can clear a 5 metres pole on the Earth. How high a pole could he clear on the Moon?
A simplified account of special relativity and the twins paradox.
A ball whooshes down a slide and hits another ball which flies off the slide horizontally as a projectile. How far does it go?
We have had two very impressive, and beautifully presented, solutions to this problems, from Laura , from The Henrietta Barnett School and Steven , from City of Sunderland College. We need the values of 1cm, 1s, 1J and 1K in terms of imperial units. These are $$ 1cm = \frac{1}{30.48}ft\quad1g = \frac{2.2}{1000}lb\quad 1J = \frac{1}{4.184} cal\quad 1K = 1.8^\circ F $$ These values can now be substituted into the numerical expressions for $G, h, c, k$. For Newton's constant we have $$ \begin{eqnarray} G& =& 6.674 \times 10^{-8} cm^3 g^{-1} s^{-2}\\ &=& 6.674 \times 10^{-8} \left(\frac{1}{30.48}ft\right)^3\left(\frac{2.2}{1000}lb\right)^{-1}s^{-2}\\ &=& \frac{6.674\times 10^{-8}\times 1000}{30.48^3\times 2.2}ft^3 lb^{-1} s^{-2}\\ &=& 1.071\times 10^{-9}ft^3lb^{-1}s^{-2} \end{eqnarray} $$ In a similar way, we can find that $$ c = 9.836\times 10^8 ft s^{-1}\quad \quad h = 2.496\times 10^{-33} lb ft^2 s^{-1} \quad\quad k = 1.83\times 10^{-24}cal ^\circ F$$ For the second part of the problem, Steven noted the following In a system of units where $G, c, h, k$ are numerically equal to 1, let $L, M, T$ be the units of length, mass and time respectively, so that $$ \begin{eqnarray} G&=& 1 L^3M^{-1}T^{-2}\\ c&=& 1LT^{-1}\\ h&=& 1ML^2 T^{-1} \end{eqnarray} $$ We now need to solve for $M, L, T$ in terms of $c, G, h$ After some rearranging, Steven found out that $$ M=\sqrt{\frac{ch}{G}}\quad\quad L=\sqrt{\frac{Gh}{c^3}}\quad\quad T=\sqrt{\frac{Gh}{c^5}} $$ Putting in the numerical values given in the question yielded $$ M = 2.176\times 10^{-5}g\quad\quad L = 1.616\times 10^{-33}cm\quad\quad T = 5.389\times 10^{-44}s $$ Steven finally concluded with a clever extension in which he found a natural temperature scale for the universe as follows: From the equation Work = Force x Distance it can be seen that $J = Nm$, and from the equation Force = Mass x Acceleration we have thatn $N = kg m s^{-2}$. So, the units of the Bolzmann constant are $kgm^2s^{-2}K^{-1}$ So, as before, supposing the $\theta$ is the natural temperature scale for the universe $$ k = 1 ML^2T^{-2}\theta^{-1} $$ Solving as before gives $$ \theta = \sqrt{\frac{c^5h}{Gk^2}} $$ The Boltzmann constant must be converted to have the same units as the other constants before evaluating this result. Putting in the numbers gives $$ k = 1.38\times10^{-16}g cm^2 s^{-2}K^{-1} $$ We can now substitute this value into the expression for $\theta$ to give $$ \theta = 1.42\times 10^{32}K $$