The change in time is given by $\Delta t = \frac{1}{2} \alpha \Delta \theta \times t$.

• $12 = \frac{1}{2} \alpha (40 - \theta) \times 86400$
• $-4 = \frac{1}{2} \alpha (20 - \theta) \times 86400$
• Dividing the equations gives $\theta = 25^{\circ}\text{C}$.
• Substituting $\theta$ back gives $\alpha = 1.85 \times 10^{-5}/^{\circ}\text{C}$.

By time-reversal symmetry: the last $t$ seconds of ascent is equivalent to free fall from rest for $t$ seconds.

Distance $= \dfrac{1}{2}gt^2$

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The roots of $x^2-x+1=0$ are $x=\frac{1\pm\sqrt{-3}}{2}=-\omega,-\omega^2$ where $\omega=e^{2\pi i/3}$.

Let $\alpha=-\omega, \beta=-\omega^2$. Then:

$\alpha^{101}=(-\omega)^{101}=-\omega^{101}=-\omega^{101\mod3}=-\omega^2$ (since $101=33\times3+2$)

$\beta^{107}=(-\omega^2)^{107}=-\omega^{214}=-\omega^{214\mod3}=-\omega^1$ (since $214=71\times3+1$)

$\alpha^{101}+\beta^{107}=-\omega^2-\omega=-(w+\omega^2)=-(-1)=\mathbf{1}$