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Hi !
It would be my pleasure. The first step is to isolate \(t\) in the equation.
$$y_f=\frac{1}{2}at^2+h$$
$$y_f-h=\frac{1}{2}at^2$$
$$2(y_f-h)=at^2$$
$$\frac{2(y_f-h)}{a}=t^2$$
$$\sqrt{\frac{2(y_f-h)}{a}}=\sqrt{t^2}$$
$$\sqrt{\frac{2(y_f-h)}{a}}=t$$
And since the objet is going from the reference \(y=0\) down to 900 mm, \(\Delta y=0m-900mm=0m-0,900m\).
I hope this helps and if you have any other questions, feel free to ask them.
Have a nice day !