How Long Will It Take A Steel Ball To Fall One Meter Starting From Rest?

[Zyzzyva]

I have no formula for this; I'm working off prior knowledge and general intuition. So far I have: 9.8 m/s^2 (acceleration in free fall)/2 = 4.9 m/s (average speed over 1 second). So in one second the ball will have fallen 4.9 m. 1 m/4.9 m = (roughly) .2 (the fraction of the total distance with which I am concerned). I think I should be able to multiply 4.9 by .2 to find the average speed over 1 meter, and thus the time it took to fall one meter (.98 s) but that doesn't make a whole lot of sense as that would give the ball only .02 seconds to fall a further 3.9 m, according to my initial calculation of 4.9 m/s. I must be doing something wrong, but I don't know what as I don't know how this is actually supposed to be done. Can someone help me?

[bibblebub]

Starting at rest with the only force being gravity the formula is simply

d = gt²/2

where d is distance, g is the acceleration due to gravity and t is time, and rearranging that one gets

t = √(2d/g)

so the time to drop 1 meter is √(2/9.8) = 0.45 seconds.

[vascop]

The way to do it using your reasoning is as follows:
After t seconds speed is 9.8 x t. So average speed over that interval is 9.8xt/2 as you say in your question (if t=1 sec then av. speed is 9.8/2, and if t=2 then av. speed =9.8 etc).
So now the average speed after travelling for t seconds is distance/time.
So after it has travelled 1 metre its average speed is 1/t.
But above we have shown that average speed is also equal to 9.8 x t/2.
So this means that 1/t=9.8 x t/2 after 1 metre.
Rearranging this gives t^2=2/9.8, so t=srt(2/9.8)=approx 0.45

The thing to realise is that the average speed over the first second is not the same as the average speed over the second second:
Av. speed over first second is 9.8/2 metres/sec as you say.
Since after 2 seconds the speed will be 9.8 x 2 then the average speed over the 2nd second is 1/2(9.8 x 2+9.8)=3 x 9.8/2 metres/sec. So the average speed of the ball over the 2nd second is 3 times the average speed over the first second.
That's why your calculation doesn't work, because you are assuming that the average speed is the same over the same period of time, and it isn't.

[vascop]

That should of course be sqrt(2/9.8) NOT srt(2/9.8)

[jake-the-peg]

There are 3 formulae that solve questions like this

v=u+at

s=ut+½at²

v²=u²+2as

v is final velocity, u initial velocity, a acceleration, t time and s distance

With these you should be able to work out any probelem with constant acceleration

[vascop]

The trouble with using formulae without understanding them is that you gain no insight into what's going on and everything remains a mystery.

[Teddio]

Acceleration is a measure of the rate of change of velocity. A falling object, neglecting air resistance, will increase its velocity by 9.8 m/s every second. Jake's formulae will allow you to arrive at the correct answer (assuming zero air resistance and uniform acceleration).

The units of acceleration make it difficult to appreciate the concept of acceleration. Consider a fast car which can get from zero to 60mph in 3 seconds. Then its acceleration will be 20mph/s. That is, its velocity will increase by 20mph every second.

[Peter Pedant]

I was gonna do Bibblebub;s.... but he got there first
short and sweet

[wildwood]

Is it true that in quantum physics, terminal velocity can never be reached.

[bibblebub]

i thought that trying to explain integral calculus in order to derive the equations from first principles might make for a rather long answer

[bibblebub]

Terminal velocity applies to an object moving in a fluid and so is only relevant to physics at a macro level (though I'm ready to be corrected on that). What you might be thinking about with quantum physics is the uncertainty principle, the result of which is that the more accurately you try to measure an particle's momentum (of which velocity is one factor) the less accurately you can ascertain its position, and vice versa.

[owdhamer]

Ask DT,or Moonie.

[Zyzzyva]

Thank you everyone! Glad to have that formula and explanation :)

[vascop]

@bibblebub
The integral calculus is not required to derive these equations.

[Hopkirk]

My brain hurts

[vascop]

@bibblebub
Terminal velocity also applies to steel balls falling through the air under gravity.

[mibn2cweus]

Since an acceleration is involved the velocity and therefore the distance covered at each time interval is changing.

More discussion of acceleration here - http://www.physicsclassroom.com/class/1dkin/u1l1e.cfm

[bibblebub]

vascop - yes, and air is a fluid (look it up in a dictionary)

[vascop]

@bibblebub
OK, fair enough. Thanks for pointing that out.