How long will it take to fall?

Mathguru tells me this is where physics questions go, so I'll post my question here.

I am having trouble understanding acceleration. Namely, how to calculate how fast an object will get to...

So here is my made up question to help me understand. Let's say (for simplicity) that gravity is equal to 10m/s^2 instead of 9.8. Now if we drop an object from 100 meters up how long (excluding air resistance) will it take to reach the ground?

Air resistance: Linear resistive force

Assume an object falling downward from rest under the influence of gravity (near the surface of the Earth) with a resistive force on it proportional to the object's velocity. The constant b (which has no specific name I know of) depends on the properties of the air and shape and size of the object. (I don't believe it depends on the mass of the object, but that also may vary with the medium the object falls through.) We may set up Newton's 2nd Law for this object and obtain the following equation for the net force:

$\displaystyle \sum F = mg - bv = ma$ (Note: I have chosen the downward direction to be posititve here.)

(For the record, if the object were moving upward, the weight and overall acceleration are still downward, but the resistive force now has a *positive *sign, due to the vector nature of v. The nature of the differential equation and solution remains the same. Note that if we throw this object upward we have to solve this equation first with the positive sign on bv until it reaches maximum height, then use the equation with the negative sign as it begins to fall.)

Normally this would be considered an equation in terms of the position y(t), but an interesting result occurs if we solve it in terms of v(t):

$\displaystyle m \frac{dv}{dt} + bv = mg$

For an object falling from rest we have v(0) = 0, so the solution becomes:

$\displaystyle v(t) = \frac{mg}{b} \left ( 1 - e^{-bt/m} \right )$

Note that for large t v approaches a constant value $\displaystyle \frac{mg}{b}$ called the "terminal velocity," $\displaystyle v_t$.

We can easily obtain the function y(t) from the expression for v(t):

$\displaystyle y(t) = \int_0^t dt' \, v(t') = \left ( \frac{mg}{b} \right ) t - g \left ( 1 - e^{-bt/m} \right )$

-Dan

Air resistance: Quadratic resistive force

Assume an object falling downward from rest under the influence of gravity (near the surface of the Earth) with a resistive force on it proportional to the square of the object's velocity. Here the resistive force is taken to be:

$\displaystyle R = \frac{1}{2}C \rho A v^2$ (Rayleigh's formula?)

where C is the "drag coefficient", $\displaystyle \rho$ is the density of the medium (in this case air), and A is (more or less) the cross-sectional area presented to the medium as the object falls. I will presume A to be constant, though obviously most objects tumble as they fall through the air. In air, most objects have a value of C around 1/2.

The equation of motion is:

$\displaystyle \sum F = mg - \frac{1}{2}C \rho A v^2 = ma$ (Again, I am assuming positive downward.)

(In this case if the object is moving upward we can no longer think of the direction of v...v^2 has no direction! However, the resistive force is *always* against the direction of motion, so we again get a positive sign as we did for the linear resistance case. I don't *think* the nature of the solution depends on the negative sign in front of v^2 (though I haven't checked it) so the same comments should apply here as they did in the linear resistance case.)

Expressed as a function v(t) this equation is nonlinear:

$\displaystyle m \frac{dv}{dt} + \left ( \frac{C \rho A}{2m} \right ) v^2 = mg$

and is difficult to solve as most nonlinear equations tend to be. A solution can be found (please don't ask me how, I had to look up the solution myself!) for an object released from rest v(0) = 0:

$\displaystyle v(t) = \sqrt{ \frac{2mg}{\rho A C} } \, tanh \left ( t \sqrt{ \frac{g \rho A C}{2m} \right )$ For those that don't know: $\displaystyle tanh(x) = \frac{sinh(x)}{cosh(x)} = \frac{(e^x - e^{-x})/2}{(e^x + e^{-x})/2} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$.

Again we see that for large t v approaches a constant value, the terminal velocity. In this case $\displaystyle v_t = \sqrt{ \frac{2mg}{C \rho A} }$.

We may obtain the equation for y(t) by integration:

$\displaystyle y(t) = \int_0^t dt' \, v(t') = \sqrt{ \frac{2mg}{ \rho A C} } \frac{ \left ( exp \left [ 2t \sqrt{ \frac{g \rho A C}{2m} } \right ] -1 \right ) t}{exp \left [ 2t \sqrt{ \frac{g \rho A C}{2m} } \right ] +1}$

-Dan

There's a short-cut to find the terminal velocity. (This can be done in the linear resistance case as well.)

We have the equation:

$\displaystyle mg - \frac{1}{2}C \rho A v^2 = ma$

from Newton's 2nd Law. If we look for a v where the acceleration is 0 (a constant velocity, of course, is represented by this) we need to solve:

$\displaystyle mg - \frac{1}{2}C \rho A v_t^2 = 0$

for v and we get the same terminal velocity as obtained above.