I must be stupid

[Anonymous]

i think its because air friction is disregarded (therefore there is no "terminal velocity", which youre talking about too ;p)

[wallace]

Oh dear. I haven't done calculus yet.
I'll read up on it and get back to you.

Ok, I'll try to explain. You know that the acceleration due to gravity is 9.8m/s/s

so a = 9.8m/s/s
a = v1-v2/s

v1-v2/s = 9.8m/s/s

you want velocity which is meters per second so we need to figure out how to get that time unit out of the denomiator. You can do this by multiplying both sides by a time unit. Since our scale is time we can simply make a new graph with the slope of 9.8, or the first antiderivative.

v = 4.9s m/s + initialvelocity

Since we are assuming initial velocity is zero velocity is simple
4.9s m/s

now v = d/s so
d/s = 4.9s m/s

we need to get rid of another s in the demoniator to find distance an object falls over a length of time. So using the same logic as above we take the first antiderivative of the new formula which comes out to be

d = 1.6ss m

in all cases s is time in seconds.

[Zack]

Yeah, I still don't get that.
Why 4.9? A line with slope of 9.8 would be y=9.8x, not 4.9x.

[Deleter]

This uses no calculus as I haven't learned it yet, rather it uses simple physics that I have learned this year. I believe it is correct, as I got the same answer as the people above who used calculus.

using:
Vi+at = Vf,

(Vi+Vf)/2 = avg V,

and

avg V * t = d

you get:

(Vi + Vf)/2 * t = d

then substituting [ Vi + at ] for [ Vf ] you get,

(Vi + Vi+at)/2 * t = d

Assuming it is dropped, Vi=0 so,

at/2 *t = d

simplified:

at^2 / 2 =d

rearranged:

t^2 = 2d/a

so,

t= SQR( 2d/a)
(this would be plus or minus, but we cant have a negative time so that is dropped)

since Vf = at (assuming Vi = 0), then we can replace t with our sqroot to get:

a * SQR( 2d/a) = Vf

simplified:

Vf = SQR (2da) or using 9.81 for a, SQR( 19.62d )

so for your problem

Vf = SQR (2*1000*9.81)=141 meters/s

[RyanKelly]

Deleter, your solution works because the equation given for the velocity is linear. This equation is actually an approximation based on observed data. Under common conditions this linear fit is usefull, and a linear equation was settled because it makes your solution possible.


By the way. I love this thread. It's great to see the same problem tackled in different ways.

[Zack]

Oy Vey. I don't understand how any of those equations are derived.

[SJ Zero]

If acceleration is constant, then the velocity increases in a linear fashion. Because of this, x will work out in a certain fashion.

I'll work it out later, I've got homework to do.

[RyanKelly]

Acceleration is only constant because is was postulated as such in the original problem. We're working with an idealized model with little reference to any experienced world. We've already disregarded air and an untold number of other possible variables.

[pr0gger]

Some simple stuff to remember about calculus:

The dervative of a function is a graph of the rate of change.

In this example, f(x) is the function, and f'(x) (read as "f prime of x") will be the derivative. (Notice the little " ' " -- it's easy to miss)

So say you had the function

f(x) = x

At all points in that function, the rate of change (the slope) is 1. So

f'(x) = 1

Similarly, if you had the function

f(x) = 3

The rate of change is 0 at all points, because that's a horizontal line. The slope is zero, so the derivative is

f'(x) = 0

You follow?

Now, if you had a more complex function, such as

f(x) = 3x^3

It gets a little more complex. Here is the rule for those equations:

IF f(x) = ax^n

THEN f'(x) = nax^(n-1)

That's called the POWER RULE. Therefore:

f(x) = 3x^3
f'(x) = (3*3)x^(3-1)
f'(x) = 9x^2

What does that mean? The slope at any given point (x) in the equation f(x) =3x^3 can be found by plugging x into the derivative f'(x) = 9x^2.

Here's another example: You have an equation f(x) = 4x^2 ; therefore, the derivative is f'(x) = 8x.
--------
Some things to remember:

A derivative can also be written as dy/dx. So if your equation is y =10x, then your derivatve could be

y' = 10

OR

dy/dx = 10

Both mean the same thing.

--------------

OK! now that all that basic stuff is out of the way, we can deal with your problem.

Your equation for velocity is

Velocity = Acceleration * Time

OR

V(t) = a * t

Or better yet, since a = 9.8,

V(t) = 9.8t

Where t is time, and a is the acceleration constant.

Now, velocity is measured in m/s -- that's the change in distance (measured in meters) over the change in time.

Therefore, that velocity equation is a derivative of the distance equation you're trying to find.

So, if we say d(t) is the function for distance at any given time, then d'(t) is the function for change of in distance at a given time-- which would be the velocity. Therefore:

d'(t) = V(t) = 9.8t

We have to find the function, d(t), knowing the derivative, d'(t). That's called integration. I won't go into detail about integrals-- but you should see the connection:

Given the Power Rule that I showed above...

IF f(x) = ax^n

THEN f'(x) = nax^(n-1)

It makes sense that

d(t) = 4.9t^2

You could conclude that derivatve of v(t) = 9.8t is equal to d'(t).

d(t) = 4.9t^2
d'(t) = (2* 4.9)t^(t - 1)
d'(t) = 9.8t^1
v(t) = d'(t) = 9.8t

Hope that helps with the calculus aspect of it.

--j_k

[Zack]

Yes it certainly does. Thanks progger! Best answer yet.
Now I'm excited to learn calculus.