# Dropping a penny from a tall buidling

If a penny is dropped from a tall building how long does it take for it to hit the ground ?

If dropping the penny from a building that is L1 feet tall takes S1 seconds I guess dropping from a building 2*L1 feet tall won't take 2*S1 seconds. Similarly a coin that is twice as heavy as the penny will take more than S1/2 seconds.

How does the height of the building, weight of the coin and acceleration due to gravity factor in the time taken (assuming there are no other variables like changes in wind speed) ? What is the right mathematical approach to solve this ?

This is not a homework question. I am just curious after seeing this question in a magazine.

• Relatively straightforward only on an airless planet. – André Nicolas Feb 21 '14 at 7:32
• I am confused. If everything else except acceleration is assumed negligible, then the mass of a falling object should not affect the time it takes to reach the ground. – Yiyuan Lee Feb 21 '14 at 7:39
• The key term is "assumed negligible." Air resistance plays a very significant role. – André Nicolas Feb 21 '14 at 7:49

The displacement (in meters) $s$ of an object after time (in seconds) $t$ is given (assuming acceleration remains constant) by :

$$s = \frac{1}{2}at^2$$

where $a$ is the constant acceleration of the object (approximately equivalent to the famous numerical value of $9.81$ $m$ $s^{-2}$ on earth's surface). This is really just integrating a constant twice with respect to $t$. We can omit the initial velocity because in your case you are merely "dropping" it from a height (i.e. initial velocity $= 0$).

Afterwards, we let the time taken for the object to be displaced by twice the distance ($2s$) be $t'$. Then we have

$$2s = \frac{1}{2}a(t')^2$$

Dividing both equations (assuming acceleration remains the same):

$$2 = \left(\frac{t'}{t}\right)^2\\ 2t^2 = (t')^2\\ t' = \sqrt{2}t$$

So we see that the time required will increase by a factor of $\sqrt2$.

• As also Americans read this thread (and in fact also in general), you should add that the $9,81$ is really $9.81\frac{\text m}{\text s^2}$ and not just a number. And the acceleration is not "initial" in the sense of "taking place only at the beginning of the fall", it accelerates all the time – Hagen von Eitzen Feb 21 '14 at 7:39
• @HagenvonEitzen Indeed. Updated, thanks! – Yiyuan Lee Feb 21 '14 at 7:41