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Suppose that there is a object that does a y-axis-only free fall to ground. The initial distance from the ground is defined as $H$.

How does one prove that time the object takes to reach the ground is $T=\sqrt{2H/g}$?

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closed as too localized by Qmechanic, Manishearth, Emilio Pisanty, Waffle's Crazy Peanut Dec 19 '12 at 9:50

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Please see our homework policy. We expect homework problems to have some effort put into them, and deal with conceptual issues. If you edit your question to explain (1) What you have tried, (2) the concept you have trouble with, and (3) your level of understanding, I'll be happy to reopen this. (Flag this message for ♦ attention with a custom message, or reply to me in the comments with @Manishearth to notify me) – Manishearth Dec 19 '12 at 9:50

freee fall equations

$$ v=gt $$

since initial velocity is 0

$$ y= \frac{1}{2}gt^{2} $$

from this secon equation we can get t as $$ t= \sqrt{\frac{2y}{g}} $$

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While Jose's answer is correct, I think it's worth showing the working to help you see exactly how this is done. When you release the object we know that it's acceleration will be $g$ (9.81m/sec). Writing this as a differential equation gives:

$$ \frac{d^2y}{dt^2} = g $$

Because $g$ is a constant, we can immediately integrate this:

$$ \frac{dy}{dt} = \int g.dt = gt + C $$

where $C$ is the constant of integration. To find $C$ we note that the speed, $dy/dt$ is zero when $t$ is zero, so that means $C = 0$. Now, to find $y$ as a function of time we integrate again:

$$ y = \int gt \space dt = \frac{1}{2}gt^2 + D $$

where again $D$ is a constant of integration. Since $y = 0$ when $t = 0$ we know $D = 0$ as well, so the equation of motion is just:

$$ y = \frac{1}{2}gt^2 $$

and to find the time taken to travel a distance $y$ we just rearrange this to give:

$$ t = \sqrt{\frac{2y}{g}} $$

Since I've gone this far, I might as well cover the case when the object isn't dropped from rest but is thrown at some velocity $u$. Our first integration gave:

$$ \frac{dy}{dt} = gt + C $$

but if the object is thrown with initial velocity $u$, when $t = 0$ $dy/dt = u$. That means the constant of integration $C = u$ and:

$$ \frac{dy}{dt} = u + gt $$

If we integrate this again we get an equation that you've probably seen in your Physics books but may have never seen derived:

$$ y = ut + \frac{1}{2}gt^2 $$

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