<s>[INST] A marble is dropped straight down from point $A$ 
A marble is dropped straight down from point $A$. The time required
  for the ball to travel the distance $h_1$ between $A$ and $B$ is equal
  to the required time to travel the distance $h_2$ between $B$ and $C$.
  Find $\dfrac{h_1}{h_2}$ by using $g \approx 10$ $m/s^2$.

I am not sure how to start the solution. All the distance is: $h=\dfrac{gt^2}{2}$, where $g\approx$ $10$ $m/s^2$ and $h=\dfrac{10(t+t)^2}{2}=\dfrac{10(2t)^2}{2}=\dfrac{10\cdot 4t^2}{2}=\dfrac{40t^2}{2}=20t^2$. We can calculate $h_1$ : 
  $h_1=\dfrac{a_1t^2}{2}$. Here ($h_2$) we have initial speed, that according to me is equal to $a_1t$. Therefore, $h_2=\dfrac{a_1t^2+a_2t^2}{2}$ and$h_1+h_2=\dfrac{a_1t^2}{2}+\dfrac{a_1t^2+a_2t^2}{2}=\dfrac{2a_1t^2+a_2t^2}{2}=20t^2$. It seems like I can't finish the problem in this way. 

 [/INST] From first principles. The equation for velocity is
$ v = u + ft $
The equation for distance is 
$s = ut + \frac{1}{2} f t^2$

So from $A$ to $B$
$v1 = gt $
$h1 = \frac{1}{2} g t^2$ 
And from $B$ to $C$
$h2 = gt^2 + \frac{1}{2} g t^2$

So 

$ \frac {h 1}{h 2} = \frac{\frac{1}{2} g t^2}{gt^2 + \frac{1}{2} g t^2}$
Cancelling out
$ \frac {h 1}{h 2} = \frac{\frac{1}{2}}{1 + \frac{1}{2}}  = \frac{1}{3}$
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