# How long will it take to fall?

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Aug 25th 2006, 02:24 PM
Quick
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?
• Aug 25th 2006, 02:52 PM
topsquark
Quote:

Originally Posted by Quick
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?

If we assume that the acceleration is constant the equation of motions are:
$\displaystyle y = y_0 + v_0t + (1/2)at^2$
$\displaystyle y = y_0 + (1/2)(v_0 + v)t$
$\displaystyle v = v_0 + at$
$\displaystyle v^2 = v_0^2 + 2a(y - y_0)$

The first thing to do is set up a coordinate system. I will choose upward as the positive direction and the ground as the origin.

So, we know that the original position y0 is at 100 m. The initial speed is (presumably) 0 m/s because we are dropping, not throwing, the object. The acceleration is (note) -10 m/s^2 since the acceleration is in the downward direction. We want t when y = 0.

The obvious equation to use, then, is the first equation.
$\displaystyle 0 = 100 + 0t + (1/2)*(-10)t^2$

or $\displaystyle t = \pm \sqrt{20}$ s. We obviously discard the negative solution.

-Dan
• Aug 25th 2006, 04:47 PM
dan
Quote:

Originally Posted by Quick
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?

I'm not sure if this is right...but i've heard that $\displaystyle t=.25sqrt(d)$
were t is seconds and d is feet.
has sombody heard of this?
by the way how do you write the code for a radical
dan
• Aug 25th 2006, 06:30 PM
Quick
thanx dan, now to make things more difficult...

Let's say that there is air resistance, and it takes the object 5 seconds to reach the ground instead of $\displaystyle \sqrt{20}$

Is there a way to calculate the Force of the air?

also, in the equation: $\displaystyle v=v_0+at$ is v equal to the ending speed or the average speed?
• Aug 26th 2006, 09:04 AM
CaptainBlack
Quote:

Originally Posted by dan
I'm not sure if this is right...but i've heard that $\displaystyle t=.25sqrt(d)$
were t is seconds and d is feet.
has sombody heard of this?
by the way how do you write the code for a radical
dan

\sqrt{}

RonL
• Aug 26th 2006, 09:12 AM
CaptainBlack
Quote:

Originally Posted by dan
I'm not sure if this is right...but i've heard that $\displaystyle t=.25sqrt(d)$
were t is seconds and d is feet.
has sombody heard of this?

In customary units $\displaystyle g=32 \mbox{ ft/s^2}$. Starting from rest at displacement $\displaystyle d=0$, we have:

$\displaystyle a=-g$,
$\displaystyle v=-gt$,
$\displaystyle d=-gt^2/2$,

so the time to fall $\displaystyle d \mbox{ ft}$ is:

$\displaystyle t=\frac{\sqrt{d}}{\sqrt{g/2}}=0.24 \sqrt{d} \mbox{ s}$.

RonL
• Aug 26th 2006, 05:34 PM
dan
nice proof capt'...thanks that never realy made sence to me before
• Aug 26th 2006, 05:45 PM
topsquark
Quote:

Originally Posted by Quick
thanx dan, now to make things more difficult...

Let's say that there is air resistance, and it takes the object 5 seconds to reach the ground instead of $\displaystyle \sqrt{20}$

Is there a way to calculate the Force of the air?

also, in the equation: $\displaystyle v=v_0+at$ is v equal to the ending speed or the average speed?

Now you're talking Newton's laws. There is a way, but you have to solve a differential equation. Typically in air resistance problems we take the resistive force to be proportional to the object's speed or the square of the object's speed. The first is easy to solve, the other a bit more complicated. As I know you don't know differential calculus I won't go any further unless someone else makes a request. (Besides, I don't do air resistance much and I'm 50 miles from my books at the moment. :) )

-Dan
• Aug 26th 2006, 05:50 PM
topsquark
Quote:

Originally Posted by topsquark
$\displaystyle y = y_0 + v_0t + (1/2)at^2$
$\displaystyle y = y_0 + (1/2)(v_0 + v)t$
$\displaystyle v = v_0 + at$
$\displaystyle v^2 = v_0^2 + 2a(y - y_0)$

I apologize for not labelling the variables in my first e-mail. I had two nieces pestering me when I wrote this, so I was rather distracted. :)

t is the time elapsed since the beginning of the problem. t0 (not in the above equations) has been set to 0 s.

y and y0 are the final and initial positions (y = y(t), y0 = y(0).)

v and v0 are the final and initial velocities.

a is the (constant) acceleration.

Since the acceleration is constant (and ONLY because it is constant!) we may say that the average velocity is:
$\displaystyle \bar{v} = (1/2)(v_0 + v)$
This statement is NOT true in general!

-Dan
• Aug 26th 2006, 06:05 PM
Quick
Quote:

Originally Posted by topsquark
Now you're talking Newton's laws. There is a way, but you have to solve a differential equation. Typically in air resistance problems we take the resistive force to be proportional to the object's speed or the square of the object's speed. The first is easy to solve, the other a bit more complicated. As I know you don't know differential calculus I won't go any further unless someone else makes a request. (Besides, I don't do air resistance much and I'm 50 miles from my books at the moment. :) )

-Dan

That's disapointing, I was considering doing my year long physics project on this...
• Aug 26th 2006, 06:13 PM
topsquark
Quote:

Originally Posted by Quick
That's disapointing, I was considering doing my year long physics project on this...

(Shrugs) You still can. Any Intro level Physics text (College level, of course) should have the solved equations. The solutions are in terms of common functions and all. You might even be able to find the solutions on (shudders) Wikipedia or something.

-Dan
• Aug 28th 2006, 01:15 PM
topsquark
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
• Aug 28th 2006, 02:06 PM
topsquark
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.
• Aug 28th 2006, 06:27 PM
Quick
Thanks dan, you get a +rep :). I have no idea what how to calculate most of the stuff but it's always nice to see what would be involved...
• Aug 28th 2006, 06:54 PM
ThePerfectHacker
There is a simple formula you can use.
If you are standing at $\displaystyle s_0$ and you throw an upward or downward (this makes it positive or negative) with accelaration of gravity $\displaystyle g$. Then the function that tells you the height after $\displaystyle t$ seconds is:
$\displaystyle S(t)=-\frac{1}{2}gt^2+v_0 t+s_0$
Of course in a vacuum.
---
Topsquark, may I ask you two questions?

It is true that when an object travels further up its effect from gravity diminishes. So what equation describes this, a cubic?
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last