How long does it take and how far does it travel a sphere in free fall before reaching a percentage of the terminal velocity?

Question

Assume a sphere (hail) that is free dropping in air. At some the sphere will reach close to the terminal velocity. The question is:

• how long $t_p$ in seconds does it take before a percentage $p$ of the terminal velocity is reached
• how far $t_p$ in m does the sphere travel before a percentage $p$ of the terminal velocity is reached

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This is an extension to a previous question.

Although I found (eventually) the equation for terminal velocity, I couldn't find the time and the distance, so I opted for deriving them. I'm putting up the extended derivation for scrutiny, and/or alternative answers.

Answer 1

I suggest using the relationship below to derive the equations you are looking for.

$F = ma$

$W = mg$

$D = \dfrac{C_d\rho V^2A}{2}$

$C_d$ = Drag Coefficient (Shape dependent)

$\rho$ = Atmospheric Density

The terminal velocity is reached when $W = D$,

$mg = \dfrac{C_d\rho V^2A}{2}$, thus

$V_t = \sqrt{\dfrac{2mg}{C_d \rho A}}$

Note, at this stage, $a = \dfrac{mg - D}{m} = 0$.

Force equilibrium needs to be maintained throughout the fall:

$F = ma = \dfrac{dV}{dt} = mg -D$

$\dfrac{dV}{dt} = g(1 - \dfrac{C_d\rho A}{2mg})V^2$

From here, you shall be able to derive the equation for "$t$" (see ref. 2 for derivation), and distance traveled "$s$" at any given time.

ADD: On approximately halfway through the linked wiki article (ref 2), at the bottom of "Derivation for Terminal Velocity", there is a boxed text that says "Derivation of the solution for the velocity v as a function of time t". To its far-right, click the link "show", the text will open to view.

References:

1. https://www.grc.nasa.gov/WWW/K-12/airplane/termv.html

2. https://en.wikipedia.org/wiki/Terminal_velocity#:~:text=Based%20on%20wind%20resistance%2C%20for%20example%2C%20the%20terminal,more%20closely%20as%20the%20terminal%20speed%20is%20approached.

Answer 2

TL;DR:

• the time required to reach a $p$ percentile of the Terminal velocity $$ t_p= \sqrt{\frac{2\rho_{sphere}r}{g \cdot C_D \rho_{air}}}arctanh\left(p \right)$$

• the distance travelled required to reach a $p$ percentile of the Terminal velocity

$$ x_p= \frac{2\rho_{sphere}r}{C_D \rho_{air}} \log \left(\cosh\left(arctanh\left(p \right)\right )\right ) $$

where:

• p: is the percentile of the terminal velocity (a number between 0 and 1)
• $C_D$: is the drag coefficient (for a sphere is 0.5)
• $\rho_{air}$ is density of the liquid the sphere is passing through (if air then 1.225 kg/m3)
• $\rho_{sphere}$ is the density of the sphere
• $g$: the acceleration of gravity
• $r$: is the radius of the sphere

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From this point on the proof of the above equations is presented.

Differential equation

Starting from:

$$m \dfrac{du}{dt} = mg - \frac{C_D}{2} \rho_{air} \cdot A \cdot u^2 $$

And assuming a sphere (Volume = $\frac{4}{3}\pi r^3$, and crosssectional area $A= \pi r^2$, the following differential equation can be written:

$$\dfrac{du}{dt} = g - \frac{C_D}{2} \frac{\rho_{air}}{\rho_{sphere}r} \cdot u^2 $$

where:

• $C_D$: is the drag coefficient (for a sphere is 0.5)
• $\rho_{air}$ is density of the liquid the sphere is passing through (if air then 1.225 kg/m3)
• $A$: cross-sectional area of the hailstone assuming its a sphere $\frac{\pi d^2}{4} =\pi r^2 $
• $V_{t}$: the terminal velocity of the sphere
• $m$: the mass of the sphere (assuming its a sphere the volume is $\frac{4}{3}\pi r^3$, and the density of the sphere is $\rho_{sphere}$)
• $g$: the acceleration of gravity
• v: the velocity of the sphere

Assuming that the sphere starts from rest then $$u(0) = 0$$

For simplicity I am replacing:

• $a_0 = g$
• $a_1 = \frac{C_D}{2} \frac{\rho_{air}}{\rho_{sphere}r}$

and the D.E. becomes

$$\dfrac{du}{dt} = a_0 - a_1 \cdot u^2 $$

Integration and solution of the DE

This is a separable d.e. therefore:

$$\frac{1}{ a_0 - a_1 \cdot u^2}\dfrac{du}{dt} = 1$$

Integrating both sides, and assuming that in time t, the velocity is u(t) $$\int_0^{u(t)}\frac{1}{ a_0 - a_1 \cdot u^2} du= \int_0 ^t 1dt$$

Because $a_0>0$ and $a_1>0$: $$\left[\frac{1}{\sqrt{a_0\cdot a_1}} arctanh\left(\sqrt{\frac{a_1}{a_0}}u \right) \right]_0^{u(t)}= \left[t\right]_0 ^t $$ $$\frac{1}{\sqrt{a_0\cdot a_1}} arctanh\left(\sqrt{\frac{a_1}{a_0}}u(t) \right)= t $$

You can solve for $u(t)$

$$ u(t)= \sqrt{\frac{a_0}{a_1}}\tanh\left(\sqrt{a_0\cdot a_1}t\right) $$

The terminal velocity is :

$$V_t = \sqrt{\frac{a_0}{a_1}} $$

time to reach a percentile of the terminal velocity

The time $t_p$ it takes to reach a percentile $p$ of the terminal velocity can be obtained by :

$$ u(t_p) = a \cdot V_t$$ $$ \sqrt{\frac{a_0}{a_1}}\tanh\left(\sqrt{a_0\cdot a_1}t_p\right)= p \cdot \sqrt{\frac{a_0}{a_1}}$$ $$ \tanh\left(\sqrt{a_0\cdot a_1}t_p\right)= p $$ $$ t_p= \frac{1}{\sqrt{a_0\cdot a_1}}arctanh\left(p \right)$$

distance required to reach a percentile of the terminal velocity

The distance $x_p$ required to reach a percentile $p$ of the terminal velocity can be obtained by:

$$ x_p= \int_0^{t_p} u(t)dt $$

$$ x_p= \int_0^{t_p} \sqrt{\frac{a_0}{a_1}}\tanh\left(\sqrt{a_0\cdot a_1}t\right) dt $$ $$ x_p= \sqrt{\frac{a_0}{a_1}} \int_0^{t_p} \tanh\left(\sqrt{a_0\cdot a_1}t\right) dt $$ $$ x_p= \sqrt{\frac{a_0}{a_1}} \left[\frac{1}{\sqrt{a_0a_1}} \log (\cosh(\sqrt{a_0a_1}t))\right]_0^{t_p} $$ $$ x_p= \frac{1}{a_1} \left[\log (\cosh(\sqrt{a_0a_1}t))\right]_0^{t_p} $$ $$ x_p= \frac{1}{a_1} \left(\log (\cosh(\sqrt{a_0a_1}t_p))-0\right) $$ $$ x_p= \frac{1}{a_1} \log \left(\cosh\left(\sqrt{a_0a_1}t_p\right )\right ) $$

Substituting $t_p$ in:

$$ x_p= \frac{1}{a_1} \log \left(\cosh\left(\sqrt{a_0a_1}\frac{1}{\sqrt{a_0\cdot a_1}}arctanh\left(p \right)\right )\right ) $$

$$ x_p= \frac{1}{a_1} \log \left(\cosh\left(arctanh\left(p \right)\right )\right ) $$