However, to his horror, he realised at the exact moment that he dropped the apple that there was a poor unfortunate man standing directly underneat the apple, and shouted down at the man to move out of the way of the doomed fruit.

But would the man hear Isaac's shout in time? Or would he suffer the same fate as our unfortunate father of modern physics?

The question is - which will reach the man first: the apple, accelerating due to gravity, or the sound of Mr Newton's shout, travelling at the constant speed of sound?

Firstly, the apple. The apple is accelerating at 9.81 ms

^{-2}, getting faster all the time. Yes, I'm ignoring terminal velocity and air resistance for now.

The formula to use for the apple is s = ut + 1/2 a t

^{2}.

s = distance travelled

u = initial velocity = 0

t = time since apple was dropped

a = acceleration due to gravity

since u = 0, (when the apple was dropped it had zero initial speed), we have a simplified formula: s = 1/2 a t

^{2}

This tells us s (how far the apple has fallen) after so much time has passed (t). We can rearrange it to tell us how long it will take the apple to fall from a building of height s. This gives us:

√(2s/a) = t

The formula for the sound of Isaac Newton's shout is simpler:

v = s / t

s = distance

v = velocity = the speed of sound, 330 ms

^{-1}

t = time since Isaac shouted

Rearranging gives us t = s / v

Now, suppose the building Isaac standing on was 300 m high, just short of the Chrysler Building's 319m. No, the buildings in Isaac Newton's time weren't this tall, but let's just suppose he had a time machine and he made the journey.

For the apple: √ (2s/a) = t

Time taken is 7.82 seconds

And for the shout,

t = s / v = 300 m / 330 ms

^{-1}= 0.909 seconds

So, the shout will reach the ground (7.82 - 0.909 = 6.91) seconds before the apple does, giving the potential victim time to move out of the way.

From here, we can go on to work out how high a building would have to be for the apple and the sound to reach the ground at the same time. From any height above this, the apple would always land before the shout (because the apple will keep accelerating - we're still ignoring terminal velocity) and no amount of shouting would save a victim from being hit on the head by the apple.

In order to do the calculation, we simply set the time for the apple and the time for the shout to be equal. This gives us:

t = √(2s/a) = s / v

√(2s/a) = s / v

2s/a = s

^{2}/ v

^{2}

2 v

^{2}/a = s

We can now put in (substitute) the values for the speed of sound (330 ms

^{-1}) and acceleration due to gravity (9.81 ms

^{-2}), and calculate the height of the building (s).

2 x 330

^{2}/ 9.81 = 22,201 metres, whiich is very, very high indeed. To put it another way:

It's just over 73,200 ft, which is 2.5 times higher than Mount Everest.

It's twice the height on airliner's cruising altitude.

Air resistance will be less of a problem to start with, but sound needs air to travel through, and the air is so thin at those heights that the sound won't travel as well or as fast. I'm not even going to discuss the lack of air pressure; lack of breathable oxygen; the temperature (frozen apples and frostbite); terminal velocity and air resistance.

When, during an A-level class, my maths teacher posed this question, I don't think she was thinking of such things either. That's the problem with maths without science - it can give you an answer that is meaningless and useless when you actually consider the real world!

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