Maths & Terminal Velocity (1 Viewer)

[wixxy2348]

Technically it's extracurricular, as it isn't a question from school, per se.
And I appreciate the beauty and elegance of a squashed meat pie.

So I must ask ye BOSers, what IS the terminal velocity of a meat pie?

By how great a margin does the terminal velocity of the said meat pie vary whenprojected with equal force from differing heights (specifically, a table compared with the top of a building; when the table = 100cm high and the building = 100m high)?

The meat pie must weigh 150g, does the terminal velocity differ when 20g of tomato sauce is added? What if it were BBQ sauce added? Would BBQ sauce affect the results due to it's far inferior flavour to tomato sauce?

What can be said for the individual who was idiotic enough to waste a perfectly good meat pie by projecting it?

We must find the answer to these age old questions.
Worked answers only.

 

[tommykins]

I'm assuming retardation is -mkv

F = mg - mkv = ma
a = g - kv
for terminal velocity, a = 0
so g = kv
k = g/v

Now,
a = g - kv
dv/dt = g-kv
dv/(g-kv) = dt
-1/k ln(g-kv) = t + c t = 0, v = 0
.'. -1/k ln(g) = c

-1/k ln (g-kv) = t - 1/k ln(g)
rearrange
t = 1/k [ln(g) - ln(g-kv)]
= 1/k ln (g/(g-kv))

And now I totally forgot what I was going to do

 

[squeenie]

Liz, we've already been through this. Twice.

Terminal velocity is "the constant maximum velocity reached by a body falling through the atmosphere under the attraction of gravity ".

I just remembered, all objects fall at the same speed, it's just that air resistance gets in the way. Either way, it wouldn't matter if you dropped it from a coffee table, or from the top of UTS Tower, since you'll always get the same outcome: people will remember you as that weirdo who wanted to know the terminal velocity of a meat pie.

 

[wixxy2348]

I'm assuming retardation is -mkv

F = mg - mkv = ma
a = g - kv
for terminal velocity, a = 0
so g = kv
k = g/v

Now,
a = g - kv
dv/dt = g-kv
dv/(g-kv) = dt
-1/k ln(g-kv) = t + c t = 0, v = 0
.'. -1/k ln(g) = c

-1/k ln (g-kv) = t - 1/k ln(g)
rearrange
t = 1/k [ln(g) - ln(g-kv)]
= 1/k ln (g/(g-kv))

And now I totally forgot what I was going to do

Oh and you were going so well!

 

[wixxy2348]

Also... it's not to be projected fast enough that it can escape the atmosphere (i.e. escape velocity), it is most definitely falling and thus affected by gravity.

 

[Forbidden.]

回复: Re: Maths & Terminal Velocity

I'm assuming retardation is -mkv

F = mg - mkv = ma
a = g - kv
for terminal velocity, a = 0
so g = kv
k = g/v

Now,
a = g - kv
dv/dt = g-kv
dv/(g-kv) = dt
-1/k ln(g-kv) = t + c t = 0, v = 0
.'. -1/k ln(g) = c

-1/k ln (g-kv) = t - 1/k ln(g)
rearrange
t = 1/k [ln(g) - ln(g-kv)]
= 1/k ln (g/(g-kv))

And now I totally forgot what I was going to do

Resistivity force is F_r = kv

Liz, we've already been through this. Twice.

Terminal velocity is "the constant maximum velocity reached by a body falling through the atmosphere under the attraction of gravity ".

I just remembered, all objects fall at the same speed, it's just that air resistance gets in the way. Either way, it wouldn't matter if you dropped it from a coffee table, or from the top of UTS Tower, since you'll always get the same outcome: people will remember you as that weirdo who wanted to know the terminal velocity of a meat pie.

Well the constant maximum speed after a while (mathematically, take the limit of the decreasing exponential function as t approaches infinity).

Oh and you were going so well!

Another way to solve that problem is the use of differential equations in which secondary students are not familiar with.
I will post my version if requested.

 

[cutemouse]

I just remembered, all objects fall at the same speed

No, they don't. So you're saying that if I drop a mass 10m above the ground, its velocity at say t=1 and t=15 will be the same?

All objects/masses accelerate towards Earth at the same rate, ignoring air resistance.

Acceleration is the rate of change of velocity (or speed).

 

[Omium]

No, they don't. So you're saying that if I drop a mass 10m above the ground, its velocity at say t=1 and t=15 will be the same?

All objects/masses accelerate towards Earth at the same rate, ignoring air resistance.

Acceleration is the rate of change of velocity (or speed).

Given enough distance through which to fall,

The terminal speeds will equal

 

[Lukybear]

Was that a trick question?

 

[cutemouse]

Given enough distance through which to fall,

The terminal speeds will equal

Yes, but that's terminal velocity.. Note what squeenie said:

I just remembered, all objects fall at the same speed, it's just that air resistance gets in the way