Is there a way to calculate the time taken by a falling object to reach terminal velocity? [duplicate]
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This question already has an answer here:
Calculating time to reach certain velocity with drag force
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I know it is possible to calculate terminal velocity using air density, mass, and drag coefficient, but is there any way to calculate the time taken until that speed is reached (assuming air density is constant for simplification purposes)?
homework-and-exercises newtonian-mechanics projectile drag
edited Nov 7 at 9:18
knzhou
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asked Nov 6 at 16:56
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Nov 9 at 5:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Calculating time to reach certain velocity with drag force
3 answers
I know it is possible to calculate terminal velocity using air density, mass, and drag coefficient, but is there any way to calculate the time taken until that speed is reached (assuming air density is constant for simplification purposes)?
homework-and-exercises newtonian-mechanics projectile drag
edited Nov 7 at 9:18
knzhou
37.5k9104182
asked Nov 6 at 16:56
phenolicdeath
646
marked as duplicate by sammy gerbil, John Rennie newtonian-mechanics
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Nov 9 at 5:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
7
The time is infinite - i.e. the falling object's velocity is never exactly as fast as the terminal velocity. If you want to know how long it takes to get to say 99% of the terminal velocity, that is a better question!
– alephzero
Nov 6 at 17:10
4
@alephzero: Well, in a more realistic scenario where the density is higher near the ground, an object falling from high enough above will eventually reach its "terminal" velocity (momentarily, relative to the current density). And then its speed will go down as the air gets denser, and the object will actually reach the ground at super-terminal velocity.
– Meni Rosenfeld
Nov 7 at 1:49
If an object has varying drag (for example is a skydiver, or is not a sphere and is tumbling), its terminal velocity will be different according to its orientation. In this scenario it may exceed its terminal velocity at some times.
– Ben
Nov 7 at 13:58
1
@Ben: Even for a sphere, the drag will not be constant because Cd typically varies with Reynolds number, which will be continually decreasing until the terminal velocity is reached.
– D. Halsey
Nov 7 at 22:44
add a comment |
up vote
11
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favorite
up vote
11
down vote
favorite
This question already has an answer here:
Calculating time to reach certain velocity with drag force
3 answers
I know it is possible to calculate terminal velocity using air density, mass, and drag coefficient, but is there any way to calculate the time taken until that speed is reached (assuming air density is constant for simplification purposes)?
homework-and-exercises newtonian-mechanics projectile drag
edited Nov 7 at 9:18
knzhou
37.5k9104182
asked Nov 6 at 16:56
phenolicdeath
646
This question already has an answer here:
Calculating time to reach certain velocity with drag force
3 answers
I know it is possible to calculate terminal velocity using air density, mass, and drag coefficient, but is there any way to calculate the time taken until that speed is reached (assuming air density is constant for simplification purposes)?
This question already has an answer here:
Calculating time to reach certain velocity with drag force
3 answers
homework-and-exercises newtonian-mechanics projectile drag
homework-and-exercises newtonian-mechanics projectile drag
edited Nov 7 at 9:18
knzhou
37.5k9104182
asked Nov 6 at 16:56
phenolicdeath
646
edited Nov 7 at 9:18
knzhou
37.5k9104182
asked Nov 6 at 16:56
phenolicdeath
646
edited Nov 7 at 9:18
knzhou
37.5k9104182
edited Nov 7 at 9:18
knzhou
37.5k9104182
edited Nov 7 at 9:18
knzhou
37.5k9104182
37.5k9104182
asked Nov 6 at 16:56
phenolicdeath
646
asked Nov 6 at 16:56
phenolicdeath
646
asked Nov 6 at 16:56
phenolicdeath
646
646
marked as duplicate by sammy gerbil, John Rennie newtonian-mechanics
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Nov 9 at 5:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Nov 9 at 5:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
7
The time is infinite - i.e. the falling object's velocity is never exactly as fast as the terminal velocity. If you want to know how long it takes to get to say 99% of the terminal velocity, that is a better question!
– alephzero
Nov 6 at 17:10
4
@alephzero: Well, in a more realistic scenario where the density is higher near the ground, an object falling from high enough above will eventually reach its "terminal" velocity (momentarily, relative to the current density). And then its speed will go down as the air gets denser, and the object will actually reach the ground at super-terminal velocity.
– Meni Rosenfeld
Nov 7 at 1:49
If an object has varying drag (for example is a skydiver, or is not a sphere and is tumbling), its terminal velocity will be different according to its orientation. In this scenario it may exceed its terminal velocity at some times.
– Ben
Nov 7 at 13:58
1
@Ben: Even for a sphere, the drag will not be constant because Cd typically varies with Reynolds number, which will be continually decreasing until the terminal velocity is reached.
– D. Halsey
Nov 7 at 22:44
add a comment |
7
The time is infinite - i.e. the falling object's velocity is never exactly as fast as the terminal velocity. If you want to know how long it takes to get to say 99% of the terminal velocity, that is a better question!
– alephzero
Nov 6 at 17:10
4
@alephzero: Well, in a more realistic scenario where the density is higher near the ground, an object falling from high enough above will eventually reach its "terminal" velocity (momentarily, relative to the current density). And then its speed will go down as the air gets denser, and the object will actually reach the ground at super-terminal velocity.
– Meni Rosenfeld
Nov 7 at 1:49
If an object has varying drag (for example is a skydiver, or is not a sphere and is tumbling), its terminal velocity will be different according to its orientation. In this scenario it may exceed its terminal velocity at some times.
– Ben
Nov 7 at 13:58
1
@Ben: Even for a sphere, the drag will not be constant because Cd typically varies with Reynolds number, which will be continually decreasing until the terminal velocity is reached.
– D. Halsey
Nov 7 at 22:44
7
7
The time is infinite - i.e. the falling object's velocity is never exactly as fast as the terminal velocity. If you want to know how long it takes to get to say 99% of the terminal velocity, that is a better question!
– alephzero
Nov 6 at 17:10
The time is infinite - i.e. the falling object's velocity is never exactly as fast as the terminal velocity. If you want to know how long it takes to get to say 99% of the terminal velocity, that is a better question!
– alephzero
Nov 6 at 17:10
4
4
@alephzero: Well, in a more realistic scenario where the density is higher near the ground, an object falling from high enough above will eventually reach its "terminal" velocity (momentarily, relative to the current density). And then its speed will go down as the air gets denser, and the object will actually reach the ground at super-terminal velocity.
– Meni Rosenfeld
Nov 7 at 1:49
@alephzero: Well, in a more realistic scenario where the density is higher near the ground, an object falling from high enough above will eventually reach its "terminal" velocity (momentarily, relative to the current density). And then its speed will go down as the air gets denser, and the object will actually reach the ground at super-terminal velocity.
– Meni Rosenfeld
Nov 7 at 1:49
If an object has varying drag (for example is a skydiver, or is not a sphere and is tumbling), its terminal velocity will be different according to its orientation. In this scenario it may exceed its terminal velocity at some times.
– Ben
Nov 7 at 13:58
If an object has varying drag (for example is a skydiver, or is not a sphere and is tumbling), its terminal velocity will be different according to its orientation. In this scenario it may exceed its terminal velocity at some times.
– Ben
Nov 7 at 13:58
1
1
@Ben: Even for a sphere, the drag will not be constant because Cd typically varies with Reynolds number, which will be continually decreasing until the terminal velocity is reached.
– D. Halsey
Nov 7 at 22:44
@Ben: Even for a sphere, the drag will not be constant because Cd typically varies with Reynolds number, which will be continually decreasing until the terminal velocity is reached.
– D. Halsey
Nov 7 at 22:44
add a comment |
1 Answer
1
active
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23
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accepted
A falling object doesn’t reach terminal velocity; it approaches terminal velocity asymptotically according to the formula
$$v=sqrt{frac{2mg}{rho A C_d}}tanh{left(tsqrt{frac{grho A C_d}{2m}}right)}.$$
Here $m$ is the mass of the object, $g$ is the acceleration due to gravity, $rho$ is the density of the fluid through which the object is falling, $A$ is the projected area of the object, and $C_d$ is the coefficient of drag.
So $$v_t=sqrt{frac{2mg}{rho A C_d}}$$ is the terminal velocity and $$tau=sqrt{frac{2m}{grho A C_d}}=frac{v_t}{g}$$ is the time scale on which the terminal velocity is approached according to $$v=v_ttanh{frac{t}{tau}}.$$ At $t=tau$ the object is at 76% of terminal velocity. At $t=2tau$ the object is at 96% of terminal velocity. At $t=3tau$ it is at 99.5% of terminal velocity.
answered Nov 6 at 17:20
G. Smith
2,311310
6
Note that $tanh x approx 1 - 2 e^{-2x}$ for large $x$, so the difference between $v$ and terminal velocity decreases approximately exponentially with time. This can be a helpful rule of thumb; if $v$ is 1% below $v_t$ at some time, and 0.5% below $v_t$ 10 seconds later, then $v$ will be 0.25% below $v_t$ 10 seconds after that.
– Michael Seifert
Nov 6 at 20:43
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
23
down vote
accepted
A falling object doesn’t reach terminal velocity; it approaches terminal velocity asymptotically according to the formula
$$v=sqrt{frac{2mg}{rho A C_d}}tanh{left(tsqrt{frac{grho A C_d}{2m}}right)}.$$
Here $m$ is the mass of the object, $g$ is the acceleration due to gravity, $rho$ is the density of the fluid through which the object is falling, $A$ is the projected area of the object, and $C_d$ is the coefficient of drag.
So $$v_t=sqrt{frac{2mg}{rho A C_d}}$$ is the terminal velocity and $$tau=sqrt{frac{2m}{grho A C_d}}=frac{v_t}{g}$$ is the time scale on which the terminal velocity is approached according to $$v=v_ttanh{frac{t}{tau}}.$$ At $t=tau$ the object is at 76% of terminal velocity. At $t=2tau$ the object is at 96% of terminal velocity. At $t=3tau$ it is at 99.5% of terminal velocity.
answered Nov 6 at 17:20
G. Smith
2,311310
6
Note that $tanh x approx 1 - 2 e^{-2x}$ for large $x$, so the difference between $v$ and terminal velocity decreases approximately exponentially with time. This can be a helpful rule of thumb; if $v$ is 1% below $v_t$ at some time, and 0.5% below $v_t$ 10 seconds later, then $v$ will be 0.25% below $v_t$ 10 seconds after that.
– Michael Seifert
Nov 6 at 20:43
add a comment |
up vote
23
down vote
accepted
A falling object doesn’t reach terminal velocity; it approaches terminal velocity asymptotically according to the formula
$$v=sqrt{frac{2mg}{rho A C_d}}tanh{left(tsqrt{frac{grho A C_d}{2m}}right)}.$$
Here $m$ is the mass of the object, $g$ is the acceleration due to gravity, $rho$ is the density of the fluid through which the object is falling, $A$ is the projected area of the object, and $C_d$ is the coefficient of drag.
So $$v_t=sqrt{frac{2mg}{rho A C_d}}$$ is the terminal velocity and $$tau=sqrt{frac{2m}{grho A C_d}}=frac{v_t}{g}$$ is the time scale on which the terminal velocity is approached according to $$v=v_ttanh{frac{t}{tau}}.$$ At $t=tau$ the object is at 76% of terminal velocity. At $t=2tau$ the object is at 96% of terminal velocity. At $t=3tau$ it is at 99.5% of terminal velocity.
answered Nov 6 at 17:20
G. Smith
2,311310
6
Note that $tanh x approx 1 - 2 e^{-2x}$ for large $x$, so the difference between $v$ and terminal velocity decreases approximately exponentially with time. This can be a helpful rule of thumb; if $v$ is 1% below $v_t$ at some time, and 0.5% below $v_t$ 10 seconds later, then $v$ will be 0.25% below $v_t$ 10 seconds after that.
– Michael Seifert
Nov 6 at 20:43
add a comment |
up vote
23
down vote
accepted
up vote
23
down vote
accepted
A falling object doesn’t reach terminal velocity; it approaches terminal velocity asymptotically according to the formula
$$v=sqrt{frac{2mg}{rho A C_d}}tanh{left(tsqrt{frac{grho A C_d}{2m}}right)}.$$
Here $m$ is the mass of the object, $g$ is the acceleration due to gravity, $rho$ is the density of the fluid through which the object is falling, $A$ is the projected area of the object, and $C_d$ is the coefficient of drag.
So $$v_t=sqrt{frac{2mg}{rho A C_d}}$$ is the terminal velocity and $$tau=sqrt{frac{2m}{grho A C_d}}=frac{v_t}{g}$$ is the time scale on which the terminal velocity is approached according to $$v=v_ttanh{frac{t}{tau}}.$$ At $t=tau$ the object is at 76% of terminal velocity. At $t=2tau$ the object is at 96% of terminal velocity. At $t=3tau$ it is at 99.5% of terminal velocity.
answered Nov 6 at 17:20
G. Smith
2,311310
A falling object doesn’t reach terminal velocity; it approaches terminal velocity asymptotically according to the formula
$$v=sqrt{frac{2mg}{rho A C_d}}tanh{left(tsqrt{frac{grho A C_d}{2m}}right)}.$$
Here $m$ is the mass of the object, $g$ is the acceleration due to gravity, $rho$ is the density of the fluid through which the object is falling, $A$ is the projected area of the object, and $C_d$ is the coefficient of drag.
So $$v_t=sqrt{frac{2mg}{rho A C_d}}$$ is the terminal velocity and $$tau=sqrt{frac{2m}{grho A C_d}}=frac{v_t}{g}$$ is the time scale on which the terminal velocity is approached according to $$v=v_ttanh{frac{t}{tau}}.$$ At $t=tau$ the object is at 76% of terminal velocity. At $t=2tau$ the object is at 96% of terminal velocity. At $t=3tau$ it is at 99.5% of terminal velocity.
answered Nov 6 at 17:20
G. Smith
2,311310
edited Nov 6 at 17:39
answered Nov 6 at 17:20
G. Smith
2,311310
answered Nov 6 at 17:20
G. Smith
2,311310
answered Nov 6 at 17:20
G. Smith
2,311310
2,311310
6
Note that $tanh x approx 1 - 2 e^{-2x}$ for large $x$, so the difference between $v$ and terminal velocity decreases approximately exponentially with time. This can be a helpful rule of thumb; if $v$ is 1% below $v_t$ at some time, and 0.5% below $v_t$ 10 seconds later, then $v$ will be 0.25% below $v_t$ 10 seconds after that.
– Michael Seifert
Nov 6 at 20:43
add a comment |
6
Note that $tanh x approx 1 - 2 e^{-2x}$ for large $x$, so the difference between $v$ and terminal velocity decreases approximately exponentially with time. This can be a helpful rule of thumb; if $v$ is 1% below $v_t$ at some time, and 0.5% below $v_t$ 10 seconds later, then $v$ will be 0.25% below $v_t$ 10 seconds after that.
– Michael Seifert
Nov 6 at 20:43
6
6
Note that $tanh x approx 1 - 2 e^{-2x}$ for large $x$, so the difference between $v$ and terminal velocity decreases approximately exponentially with time. This can be a helpful rule of thumb; if $v$ is 1% below $v_t$ at some time, and 0.5% below $v_t$ 10 seconds later, then $v$ will be 0.25% below $v_t$ 10 seconds after that.
– Michael Seifert
Nov 6 at 20:43
Note that $tanh x approx 1 - 2 e^{-2x}$ for large $x$, so the difference between $v$ and terminal velocity decreases approximately exponentially with time. This can be a helpful rule of thumb; if $v$ is 1% below $v_t$ at some time, and 0.5% below $v_t$ 10 seconds later, then $v$ will be 0.25% below $v_t$ 10 seconds after that.
– Michael Seifert
Nov 6 at 20:43
add a comment |
7
The time is infinite - i.e. the falling object's velocity is never exactly as fast as the terminal velocity. If you want to know how long it takes to get to say 99% of the terminal velocity, that is a better question!
– alephzero
Nov 6 at 17:10
4
@alephzero: Well, in a more realistic scenario where the density is higher near the ground, an object falling from high enough above will eventually reach its "terminal" velocity (momentarily, relative to the current density). And then its speed will go down as the air gets denser, and the object will actually reach the ground at super-terminal velocity.
– Meni Rosenfeld
Nov 7 at 1:49
If an object has varying drag (for example is a skydiver, or is not a sphere and is tumbling), its terminal velocity will be different according to its orientation. In this scenario it may exceed its terminal velocity at some times.
– Ben
Nov 7 at 13:58
1
@Ben: Even for a sphere, the drag will not be constant because Cd typically varies with Reynolds number, which will be continually decreasing until the terminal velocity is reached.
– D. Halsey
Nov 7 at 22:44