# Question #fc71d

May 9, 2017

Using Equation of motion: $s = u t + \frac{1}{2} a {t}^{2}$, here as per the drawing, $u = 0$ and $a = g$, so that: $s \propto {t}^{2} q \quad \triangle$

$\triangle \implies \frac{\frac{2}{3} h}{h} = \frac{{t}_{1}^{2}}{{t}_{2}^{2}} \implies {t}_{2}^{2} = \frac{3}{2} {t}_{1}^{2}$

In addition: ${t}_{2} - {t}_{1} = 1$

$\implies {t}_{2}^{2} = \frac{3}{2} {\left({t}_{2} - 1\right)}^{2}$

That is a quadratic that solves as ${t}_{2} = 3 \pm \sqrt{6}$ or ${t}_{2} \approx 0.55 , 5.45 \text{ sec}$

Clearly the first solution has no practical meaning [it is looking at possible solutions in negative time, which is just a mathematical consequence of the quadratic], so:

${t}_{2} = 3 + \sqrt{6} \approx 5.45 \text{ sec}$