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PeacefulPumpkin8091

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Can you show me how they got to this answer. It's been too long since I had to change exponents.

Thank you,

Ramzi Z

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 !