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is there a way to calculate time taken by a falling object to reach terminal velocity
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I know it is possible to calculate terminal velcoty using air density, mass, and drag coefficient, but is there any way to calculate time taken until that speed is reached (assuming air density is contstant for simplification purposes)
newtonian-mechanics classical-mechanics aerodynamics free-fall
asked 1 hour ago
phenolicdeath
84
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up vote
1
down vote
favorite
I know it is possible to calculate terminal velcoty using air density, mass, and drag coefficient, but is there any way to calculate time taken until that speed is reached (assuming air density is contstant for simplification purposes)
newtonian-mechanics classical-mechanics aerodynamics free-fall
asked 1 hour ago
phenolicdeath
84
3
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
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I know it is possible to calculate terminal velcoty using air density, mass, and drag coefficient, but is there any way to calculate time taken until that speed is reached (assuming air density is contstant for simplification purposes)
newtonian-mechanics classical-mechanics aerodynamics free-fall
asked 1 hour ago
phenolicdeath
84
I know it is possible to calculate terminal velcoty using air density, mass, and drag coefficient, but is there any way to calculate time taken until that speed is reached (assuming air density is contstant for simplification purposes)
newtonian-mechanics classical-mechanics aerodynamics free-fall
newtonian-mechanics classical-mechanics aerodynamics free-fall
asked 1 hour ago
phenolicdeath
84
asked 1 hour ago
phenolicdeath
84
asked 1 hour ago
phenolicdeath
84
asked 1 hour ago
phenolicdeath
84
asked 1 hour ago
phenolicdeath
84
84
3
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
1 hour ago
add a comment |Â
3
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
1 hour ago
3
3
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
1 hour ago
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
1 hour ago
add a comment |Â
1 Answer
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A falling object doesn’t reach terminal velocity; it approaches terminal velocity asymptotically according to the formula
$$v=sqrtfrac2mgrho A C_dtanhleft(tsqrtfracgrho A C_d2mright).$$
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 $$v_t=sqrtfrac2mgrho A C_d$$ is the terminal velocity and $$tau=sqrtfrac2mgrho A C_d=fracv_tg$$ is the time scale on which the terminal velocity is approached according to $$v=v_ttanhfracttau.$$ 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 1 hour ago
G. Smith
8126
Thank you very much!
– phenolicdeath
38 mins ago
And thank you for marking the answer as accepted! Lots of people posting questions forget to do that,
– G. Smith
26 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
A falling object doesn’t reach terminal velocity; it approaches terminal velocity asymptotically according to the formula
$$v=sqrtfrac2mgrho A C_dtanhleft(tsqrtfracgrho A C_d2mright).$$
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 $$v_t=sqrtfrac2mgrho A C_d$$ is the terminal velocity and $$tau=sqrtfrac2mgrho A C_d=fracv_tg$$ is the time scale on which the terminal velocity is approached according to $$v=v_ttanhfracttau.$$ 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 1 hour ago
G. Smith
8126
Thank you very much!
– phenolicdeath
38 mins ago
And thank you for marking the answer as accepted! Lots of people posting questions forget to do that,
– G. Smith
26 mins ago
add a comment |Â
up vote
5
down vote
accepted
A falling object doesn’t reach terminal velocity; it approaches terminal velocity asymptotically according to the formula
$$v=sqrtfrac2mgrho A C_dtanhleft(tsqrtfracgrho A C_d2mright).$$
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 $$v_t=sqrtfrac2mgrho A C_d$$ is the terminal velocity and $$tau=sqrtfrac2mgrho A C_d=fracv_tg$$ is the time scale on which the terminal velocity is approached according to $$v=v_ttanhfracttau.$$ 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 1 hour ago
G. Smith
8126
Thank you very much!
– phenolicdeath
38 mins ago
And thank you for marking the answer as accepted! Lots of people posting questions forget to do that,
– G. Smith
26 mins ago
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
A falling object doesn’t reach terminal velocity; it approaches terminal velocity asymptotically according to the formula
$$v=sqrtfrac2mgrho A C_dtanhleft(tsqrtfracgrho A C_d2mright).$$
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 $$v_t=sqrtfrac2mgrho A C_d$$ is the terminal velocity and $$tau=sqrtfrac2mgrho A C_d=fracv_tg$$ is the time scale on which the terminal velocity is approached according to $$v=v_ttanhfracttau.$$ 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 1 hour ago
G. Smith
8126
A falling object doesn’t reach terminal velocity; it approaches terminal velocity asymptotically according to the formula
$$v=sqrtfrac2mgrho A C_dtanhleft(tsqrtfracgrho A C_d2mright).$$
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 $$v_t=sqrtfrac2mgrho A C_d$$ is the terminal velocity and $$tau=sqrtfrac2mgrho A C_d=fracv_tg$$ is the time scale on which the terminal velocity is approached according to $$v=v_ttanhfracttau.$$ 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 1 hour ago
G. Smith
8126
edited 1 hour ago
answered 1 hour ago
G. Smith
8126
answered 1 hour ago
G. Smith
8126
answered 1 hour ago
G. Smith
8126
8126
Thank you very much!
– phenolicdeath
38 mins ago
And thank you for marking the answer as accepted! Lots of people posting questions forget to do that,
– G. Smith
26 mins ago
add a comment |Â
Thank you very much!
– phenolicdeath
38 mins ago
And thank you for marking the answer as accepted! Lots of people posting questions forget to do that,
– G. Smith
26 mins ago
Thank you very much!
– phenolicdeath
38 mins ago
Thank you very much!
– phenolicdeath
38 mins ago
And thank you for marking the answer as accepted! Lots of people posting questions forget to do that,
– G. Smith
26 mins ago
And thank you for marking the answer as accepted! Lots of people posting questions forget to do that,
– G. Smith
26 mins ago
add a comment |Â
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3
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
1 hour ago