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I know the terminal velocity equation as:

V^2=(2mg)/((CdAp)

I also know that v^2 = u^2 + 2as. Assuming the object's terminal velocity is also its final velocity, and knowing every other variable in this equation except for mass, m, can I equate the two and get an equation for mass at which an object will reach terminal velocity for a certain height, density, drag coefficient? (initial velocity is 0 and the object is dropped strictly downwards with no added forces except for gravity).

i.e., I get (after cancellation):

m = (Cd)Aps

This is the basis of one of my projects (I'm a high school student), and I want to make sure this is correct. Please let me know.

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The equation $v^2=u^2+2as$ only applies to an object that is accelerating at a constant acceleration $a$. Once you take into account drag, the acceleration of a falling object is not constant - it starts at $g$ when $t=v=0$, but then decrease asymptotically towards $0$. So you cannot use $v^2=u^2+2as$.

A falling object with drag never actually reaches its terminal velocity $V_t$ - it approaches it asymptotically (in the same way as its acceleration approaches but never reaches $0$). Its velocity at time $t$ is actually given by

$\displaystyle v(t) = V_t \tanh \left( t \frac {g} {V_t} \right)$

We can re-write this as

$\displaystyle v(t) = V_t \tanh \left( \frac t T \right)$

where

$\displaystyle T = \frac {V_t}{g} = \sqrt {\frac {2m}{g \rho A C_d}}$

$\tanh(1) \approx 0.762$, so at time $t=T$ the falling object will have reached $76.2\%$ of its terminal velocity; at time $t=2T$ it will have reached $96.4\%$ of its terminal velocity; and at time $t=3T$ it will have reached $99.5\%$ of its terminal velocity.

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  • $\begingroup$ Hey, thanks for this, but can you please let me know a little about how you got to the hyperbolic tangent equation? I don't need to know every step, but at least the general premise. $\endgroup$ Feb 19 at 7:38
  • $\begingroup$ @EmilJohnson See en.wikipedia.org/wiki/Terminal_velocity $\endgroup$
    – gandalf61
    Feb 19 at 9:06
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I also know that $v^2 = u^2 + 2as$.

Unfortunately that reasoning is incorrect because here $v$ changes instantaneously with time.

The Newtonian Equation of Motion of the falling object is:

$$ma=mg-\frac12 \rho C_DAv^2$$ Or: $$\dot{v}=g-\frac{1}{2m} \rho C_DAv^2\tag{1}$$

This differential equation shows that as the object's velocity increases, its acceleration $a=\dot{v}$ gradually decreases, until it becomes $0$ when terminal velocity is reached.

@gandalf61 has provided solutions to $(1)$.

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