What is atmospheric drag?

The gravity vs drag problem revisited with the correct answer to the question "At what time will terminal velocity be reached?"

  • Force due to gravity is mg Force due to drag is 0.5*A*p*C*v^2 where p = density of the fluid (air) A = cross-sectional area of the falling mass that is exposed to the resisting material C = drag coefficient (dependent on many factors) Terminal velocity is when the object stops accelerating and the velocity remains constant. Based on free fall equations and the drag equation, I'm wondering if terminal velocity can actually be achieved...and if so...after how long?/how far of a drop? Step 1: Let Fd = kv^2 where k is the constants mentioned above Step 2: Set Fg - Fd = Fr where Fr is the resultant force. This is rewritten as m*a(t) = m*g - k*v(t)^2 or dv/dt = g - (k/m)*v(t)^2. "a" and "v" are written as functions of time "t"/ Step 3: Solve the equation (which I did not do before) You get sqrt(m/(4gk))*[ln(1+v*sqrt(k/(gm))) - ln(1-v*sqrt(k/(gm)))] = t Step 4: Terminal velocity happens when Fg = Fd. Quickly solving, you get Vterminal = sqrt(mg/k) Step 5: Find what time this happens by plugging Vterminal in to the long equation above. It turns out that terminal velocity occurs at time infinity, which means terminal velocity can only be approached and not achieved. DISCUSS!! Read more: Object falls through air, providing resistance. The drag force is F=Z*v^2 where z is a constant. When do we achieve terminal velocity? | Answerbag http://www.answerbag.com/q_view/2165096#ixzz0sT32c3X2

  • Answer:

    In the real world, air moves around and the behavior of falling objects is a little bit chaotic since no object has a perfect, flawless surface. An object falling slower than the speed of sound has a shock wave preceding it that makes the behavior of the air molecules a little more chaotic... The actual terminal velocity will vary with the orientation of the object in the air and the local behavior of the air... the larger the object the more these small, local effects can be averaged/smeared across the problem. The actual speed of real objects falling in the atmosphere will vary a lot more than the theoretical gap between the point on the graph and the theoretical limit after a relatively short time falling.

WarHorseLeBron at Answerbag.com Visit the source

Was this solution helpful to you?

Just Added Q & A:

Find solution

For every problem there is a solution! Proved by Solucija.

  • Got an issue and looking for advice?

  • Ask Solucija to search every corner of the Web for help.

  • Get workable solutions and helpful tips in a moment.

Just ask Solucija about an issue you face and immediately get a list of ready solutions, answers and tips from other Internet users. We always provide the most suitable and complete answer to your question at the top, along with a few good alternatives below.