$A v=a v_1$

$A\left(-\frac{d y}{d t}\right)=a \sqrt{2 g y} ; d t=\frac{A}{a \sqrt{2 g}} \cdot \frac{-d y}{\sqrt{y}}$

$\int_0^{t_1} d t=\frac{A}{a \sqrt{2 g}} \int_{ h }^0-\frac{d y}{\sqrt{y}}$

$t_1=\frac{A}{a \sqrt{2 g}} 2 \sqrt{ h } ; t_1=\frac{A}{a} \sqrt{\frac{2 h}{g}}$

$A v^{\prime}=a v_2$

(image)

$A\left(-\frac{d y}{d t}\right)=a \sqrt{2 g(H+y)}$

$d t=-\frac{A}{a \sqrt{2 g}} \frac{d y}{\sqrt{H+y}}$

$\int_0^{t_2} dt =-\frac{A}{a \sqrt{2 g}} \int_H^0 \frac{d y}{\sqrt{H+y}}$

$t_2=\frac{A}{a \sqrt{2 g}}(2)(\sqrt{H+ h }-\sqrt{ H }) \quad \& H =\frac{16 h }{9}$

$=\frac{ A }{ a } \sqrt{\frac{2 h }{ g }}\left(\frac{5}{3}-\frac{4}{3}\right)$

$t _2=\frac{ A }{ a } \sqrt{\frac{2 h }{ g }}\left(\frac{1}{3}\right)$

$\text { ratio } \frac{ t _1}{ t _2}=3$