# Maximum height for Terminal Velocity to be reached for a certain mass?

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.

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.

• 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. Feb 19 at 7:38
• @EmilJohnson See en.wikipedia.org/wiki/Terminal_velocity Feb 19 at 9:06

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 $$(1)$$.