Are you sure that the drag isn't - 1.5 v PER UNIT MASS?
If that were so, the answer can be found without reference to the mass itself. Note also that in that case the 1.5 must have units of T^(-1), and that this implies that the limiting velocity is approached on timescales of a second or so!
I will address that case; you can make the necessary modifications if that isn't good enough. I find that:
HE TAKES 466.5... SECONDS TO FALL 10,000FT,
OR 7 MINUTES and 46.5... SECONDS!
Measure x DOWNWARDS. Then the basic equation of motion is:
m d^2x/dt^2 = mg - 1.5mdx/dt, or m d^2x/dt^2 + 1.5mdx/dt = mg, or:
d2x/dt^2 + 1.5 dx/dt = g.
Integrate this wrt t, to satisfy x = dx/dt = 0 at t = 0. Then:
dx/dt + 1.5 x = g t. Let c = 1/5, for simplicity, then
dx/dt + c x = g t.
There's an integrating factor e^(ct) for this, giving us
d/dt (x e^ct) = g*integral (from 0 to t) of [t e^(ct) dt].
The RHS can be integrated by parts, and the initial conditions satisfied, to give:
x e^(ct) = g/c^2 * e^(ct) (ct - 1) - g/c^2 * 1 * (- 1), that is, in the end,
x = g/c^2 * (ct - 1) + g/c^2 * e(- ct).
THAT IS THE FINAL x(t) RELATIONSHIP!
After all this nasty work, a small check is desirable. Please mail me the check at ... Oh, I'm sorry. I meant a check of the mathematics:
Does this look like free fall near t = 0?!
Well, for small t, x = g/c^2 [(ct - 1) + 1 - ct + 1/2 c^2 t^2 - 1/6 c^3 t^3 + ..]
= g/c^2 [1/2 c^2 t^2 - O(t^3)] = 1/2 g t^2 - O(t^3), as it should. (GLORY BE!)
In order to solve the problem of HOW LONG he takes to fall 10,000 ft, you now have to plug x = 10,000ft, g = 32.2 ft/s^2, and c = 1.5 into this and then SOLVE the transcendental equation:
x = g/c^2 * (ct - 1) + g/c^2 * e^(- ct), or
x = g/c^2 [(ct - 1) + e^(- ct)]
for t.
Doing THAT is far beyond my pay grade, and I've done more than enough to earn my 10 points. So, I suggest that YOU solve it, for example by some Newton Raphson scheme.
Well, maybe NOT! The time to fall will probably be so large that the exponential term can be ignored. In that case,
10,000 = 32.2/(1.5)^2 * (1.5 t - 1), or
10,000 = 14.3111... * (1.5 t - 1), so that (1.5 t - 1) = 698.757764...,
therefore t = 699.757764/1.5 = 466.505... s.
Now while (1.5 t - 1) = 698.75... , the exponential term is, in comparison, e^(- 699.75...), which is smaller than 10^(-300) and thus UTTERLY NEGLIGIBLE!
What that indicates, of course, is that for MOST of the time he was falling at the limiting velocity. However, it was necessary to solve the D.E. in order to find the constant term subtracted off the ct term in the asymptotic solution
x = g/c^2 * (ct - 1)
So in fact the solution is given by the leading terms without the exponential, and therefore:
HE TAKES 466.5... s TO FALL 10,000FT!
[NOTE that if the drag WAS - 1.5 v, INDEPENDENT of the mass involved, then c will be 1.5 / m, and you WON'T be able to get a definite numerical solution without picking a value for m. This is the problem I foresaw at the very start of the problem. But there will be one main difference in this solution. Having c = 1.5 / m for a reasonable value of m will certainly extend the time before he reaches the limiting velocity; however, by the same token it COULD make the solution less straightforward, since there could be values of m for which the exponential term might cease to be negligible.]
at terminal velocity the force = 0 so v(terminal)=mg/1.5
from this you can calculate the approximate time, (what are the units of the drag???) Calculate the distance needed to fall, without drag, to reach vo and see if it <<10,000
OR
F=dv/dt so dv/dt=mg-1.5v
This is a first order equation that has a solution for v(t) but I have forgotten how. to get it.
assume g = 10 m/sec^2. permit s = s(t) be the region of the piano at time t seconds. permit H meters be the altitude of the piano at time t=0 seconds. permit T = 6.32456 seconds be the time it took the piano to hit the floor. s' '(t) = -g s'(t) = -gt (we assume s'(0) = 0 ft/sec.) s(t) = (-g/2)t^2 + H s(T) = 0 (-g/2)T^2 + H = 0 H = (g/2)T^2 = 5 * 6.32456^2 = 200 m. the fee of the piano while it hit the floor is s'(6.32456) = -gt = -sixty 3.2456 m/sec
Comments
Are you sure that the drag isn't - 1.5 v PER UNIT MASS?
If that were so, the answer can be found without reference to the mass itself. Note also that in that case the 1.5 must have units of T^(-1), and that this implies that the limiting velocity is approached on timescales of a second or so!
I will address that case; you can make the necessary modifications if that isn't good enough. I find that:
HE TAKES 466.5... SECONDS TO FALL 10,000FT,
OR 7 MINUTES and 46.5... SECONDS!
Measure x DOWNWARDS. Then the basic equation of motion is:
m d^2x/dt^2 = mg - 1.5mdx/dt, or m d^2x/dt^2 + 1.5mdx/dt = mg, or:
d2x/dt^2 + 1.5 dx/dt = g.
Integrate this wrt t, to satisfy x = dx/dt = 0 at t = 0. Then:
dx/dt + 1.5 x = g t. Let c = 1/5, for simplicity, then
dx/dt + c x = g t.
There's an integrating factor e^(ct) for this, giving us
d/dt (x e^ct) = g*integral (from 0 to t) of [t e^(ct) dt].
The RHS can be integrated by parts, and the initial conditions satisfied, to give:
x e^(ct) = g/c^2 * e^(ct) (ct - 1) - g/c^2 * 1 * (- 1), that is, in the end,
x = g/c^2 * (ct - 1) + g/c^2 * e(- ct).
THAT IS THE FINAL x(t) RELATIONSHIP!
After all this nasty work, a small check is desirable. Please mail me the check at ... Oh, I'm sorry. I meant a check of the mathematics:
Does this look like free fall near t = 0?!
Well, for small t, x = g/c^2 [(ct - 1) + 1 - ct + 1/2 c^2 t^2 - 1/6 c^3 t^3 + ..]
= g/c^2 [1/2 c^2 t^2 - O(t^3)] = 1/2 g t^2 - O(t^3), as it should. (GLORY BE!)
In order to solve the problem of HOW LONG he takes to fall 10,000 ft, you now have to plug x = 10,000ft, g = 32.2 ft/s^2, and c = 1.5 into this and then SOLVE the transcendental equation:
x = g/c^2 * (ct - 1) + g/c^2 * e^(- ct), or
x = g/c^2 [(ct - 1) + e^(- ct)]
for t.
Doing THAT is far beyond my pay grade, and I've done more than enough to earn my 10 points. So, I suggest that YOU solve it, for example by some Newton Raphson scheme.
Well, maybe NOT! The time to fall will probably be so large that the exponential term can be ignored. In that case,
10,000 = 32.2/(1.5)^2 * (1.5 t - 1), or
10,000 = 14.3111... * (1.5 t - 1), so that (1.5 t - 1) = 698.757764...,
therefore t = 699.757764/1.5 = 466.505... s.
Now while (1.5 t - 1) = 698.75... , the exponential term is, in comparison, e^(- 699.75...), which is smaller than 10^(-300) and thus UTTERLY NEGLIGIBLE!
What that indicates, of course, is that for MOST of the time he was falling at the limiting velocity. However, it was necessary to solve the D.E. in order to find the constant term subtracted off the ct term in the asymptotic solution
x = g/c^2 * (ct - 1)
So in fact the solution is given by the leading terms without the exponential, and therefore:
HE TAKES 466.5... s TO FALL 10,000FT!
[NOTE that if the drag WAS - 1.5 v, INDEPENDENT of the mass involved, then c will be 1.5 / m, and you WON'T be able to get a definite numerical solution without picking a value for m. This is the problem I foresaw at the very start of the problem. But there will be one main difference in this solution. Having c = 1.5 / m for a reasonable value of m will certainly extend the time before he reaches the limiting velocity; however, by the same token it COULD make the solution less straightforward, since there could be values of m for which the exponential term might cease to be negligible.]
Live long and prosper.
F=mg-1,5v
at terminal velocity the force = 0 so v(terminal)=mg/1.5
from this you can calculate the approximate time, (what are the units of the drag???) Calculate the distance needed to fall, without drag, to reach vo and see if it <<10,000
OR
F=dv/dt so dv/dt=mg-1.5v
This is a first order equation that has a solution for v(t) but I have forgotten how. to get it.
I hope this helps.
Since you could not understand therein answers
http://answers.yahoo.com/question/index;_ylt=As0QB...
I’m not sure colleging is a good business for you; yet I hope a friend of yours should explain my answer;
♠ according to 2nd Newton’s law force on poor man is F=m*x’’,
where m is his mass in kg, x’’ is his acceleration (God help him!),
F=w+F2, w=mg is his weight in Newtons, F2=-p*x’ is drag force, p=1.5 kg/s,
x is distance from initial altitude h=10000ft=3048m; thus we get
♣ x’’ =g -(p/m)*x’, thence x’ +(p/m)*x =gt;
this is typical Bernoulli eqn; thus
x= g(m/p)^2 *{pt/m –(1 -exp(-pt/m))} meters;
♥ supposing his mass 75 kg, remembering g=9.8m/s^2, we get
x= 9.8*2500*{0.02t –(1 -exp(-0.02t))}, hence t=27.20 s;
x’ = 205.6 m/s, while his max speed should be 490 m/s;
click me if whatever; you are welcome;
assume g = 10 m/sec^2. permit s = s(t) be the region of the piano at time t seconds. permit H meters be the altitude of the piano at time t=0 seconds. permit T = 6.32456 seconds be the time it took the piano to hit the floor. s' '(t) = -g s'(t) = -gt (we assume s'(0) = 0 ft/sec.) s(t) = (-g/2)t^2 + H s(T) = 0 (-g/2)T^2 + H = 0 H = (g/2)T^2 = 5 * 6.32456^2 = 200 m. the fee of the piano while it hit the floor is s'(6.32456) = -gt = -sixty 3.2456 m/sec