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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

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2 Answers 2

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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