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When an object is dropped, does it accelerate to terminal velocity? or is it at terminal velocity immediately?

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  • $\begingroup$ Nothing is immediate in real life. $\endgroup$ – Steeven Nov 20 '18 at 20:15
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Conceptually, it depends on the type of object that is dropped. If the object is highly affected by air (wind) resistance ( like a feather or tissue) , terminal velocity will occur almost immediately after it is dropped. Acceleration due to freefall is still 9.8 m/s2 in any case, but air resistance will kick in and terminal velocity will occur amost immediately. For objects with almost no air resistance, terminal velocity will occur immediately before it reaches the surface (or ground) onto which is was dropped as the velocity and kinetic energy will increase due to freefall. For all objects that have some weight and are affected by air resistance (as air resistance increases as velocity increases), like skydivers, terminal velocity will occur when the force of the air resistance equals the force of gravity and acceleration will reduce to zero. At that point the body will fall at constant speed to the ground. That constant speed, by the way, may not be a slow speed and that is why there are parachutes for skydivers. Parachutes provide extra air resistance, and the skydiver will reach a new slower terminal velocity using the same reasoning as just mentioned.

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It accelerates.

Gravity is a constant force, whereas wind resistance varies with the square of velocity. So at first the wind resistance is zero, and you accelerate due to gravity alone. After time T you have accelerated to speed S, and now there is some wind resistance and you are not accelerating as fast. Eventually the drag due to speed S is the same as the force of gravity, and presto, terminal.

When I skydived, the rule of thumb was that you reached terminal after about 12 seconds, at which point you were going 120 mph.

Mathematically, the force of gravity is Fg=mg, where m is the mass and g is acceleration of gravity in whatever units you prefer. Air drag is Fd = (CdA)(V2rho)/2, where Cd is a constant you look up based on the shape, A is the area and rho is the density of air. rho A and Cd are all constants, so it really comes down to V2/2.

You're at terminal when the force of gravity is the same as the force of drag, or Fg = -Fc. Solving for V you get V = root(2mg/CdArho)

Rho for sea level is around 0.00238, and Cd is 1 for a sphere. So if you drop a sphere that's 1 square foot and weights 10 pounds, terminal is about 62 mph.

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  • $\begingroup$ Worth mentioning that mathematically you approach, but never reach, terminal velocity. $\endgroup$ – JMac Nov 20 '18 at 20:29

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