How Does Air Resistance Affect the Distance of Free Fall?

[eku_girl83]

Here's the problem:
Consider a particle of mass m whose motion starts from rest in a constant gravitational field. If a resisting force proportional to the square of the velocity (kmv2) is encountered, find the distance s the particle falls in accelerating from v0 to v1.

I began with the equation m (dv/dt)=-mg-kmv2.
From this, we see that dv = -g-kv2 dt.
When I separate variables and integrate this however, I get a function that involves the inverse tangent. I would have to integrate a second time to find distance, correct?
The book gives the distance s from v0 to v1 as 1/2k ln ((g-kv02)/(g-kv12)). What am I doing wrong?

Any help greatly appreciated!

 

[sqrt(-1)]

Use the relation

$$\ddot{x} = v \frac{dv}{dx}$$

on the left hand side of your equation and you should find things work out :)

 

[quicknote]

I would like to clarify that the equation m (dv/dt)=-mg-kmv2 is known as the differential equation for free fall with resistance. This equation represents the balance between the gravitational force (mg) and the resisting force (kmv2), where k is a constant factor and v is the velocity of the particle.

To solve this problem, we need to use calculus to find the velocity and distance of the particle as it falls. Integrating the equation, we get dv/(-g-kv2) = dt. By using the substitution u = v2, we can simplify the equation to 1/u du = (-1/gk) dt. Integrating both sides, we get ln(u) = (-t/gk) + C, where C is the constant of integration. Substituting back for v, we get ln(v2) = (-t/gk) + C.

Now, we can use the initial conditions of the problem (starting from rest, v0 = 0) to solve for the constant C. When t = 0, v = v0 = 0, which means ln(v02) = 0. Solving for C, we get C = 0.

Substituting this back into our equation, we get ln(v2) = (-t/gk). To find the distance, we need to integrate v with respect to time. We can do this by using the substitution u = v2, as we did before. Integrating both sides, we get s = (-1/gk) ln(v2) + D, where D is the constant of integration.

Using the initial conditions again, when t = 0, v = v0 = 0, which means s = 0. Solving for D, we get D = 0.

Therefore, the distance s from v0 to v1 is given by s = (-1/gk) ln(v12). This matches the answer given in the book, 1/2k ln ((g-kv02)/(g-kv12)). So, it seems like there may have been a mistake in your integration or substitution.

I would suggest double-checking your work and making sure all the steps are correct. If you are still having trouble, you can consult with a math or physics tutor for additional help. Remember, it is important to always check your work and make sure your answer makes sense in the