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

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

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.

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:

• I am glad that you validated that my methodology and equations I have used. However, I couldn't find any derivation for $t$ or $s$ in ref.2.
– NMech
Aug 18, 2021 at 13:25
• @NMch Good question, it is tricky to view the derivation - On approximately halfway through the linked wiki article, 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.
– r13
Aug 18, 2021 at 17:09

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

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:

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

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

• What happens if a1 = 0?
– r13
Aug 17, 2021 at 16:02
• for $a1$ to be zero, the drag needs to be zero. In that case the body will accelerate for even, therefore the terminal velocity is not bounded neither the time.
– NMech
Aug 17, 2021 at 17:00
• I see. By definition, the terminal velocity occurs when Weight = Drag, so a1 will not be zero. Thanks.
– r13
Aug 17, 2021 at 18:12
• thanks for the clear unswer. Aug 17, 2021 at 18:45