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The terminal velocity of an object falling towards the ground, in non-vacuum, is the speed at which the gravitational force pulling it downwards is equal and opposite to the atmospheric drag (also called air resistance) pushing it upwards. At this speed, the object ceases to accelerate downwards and falls at constant speed. An object moving downwards without power at greater than the terminal velocity (for example because it previously used power to descend, it fell from a thinner part of the atmosphere or it changed shape) will slow down until it reaches terminal velocity.
For example, the terminal velocity of a skydiver in a normal free-fall position with a closed parachute is about 195 km/h (120 Mph). It would take about 5.5 seconds to reach that speed. This speed increases to about 320 km/h (200 Mph) if the skydiver pulls in his limbs—see also freeflying. This is also the terminal velocity of the Peregrine Falcon diving down on its prey.
The reason an object reaches a terminal velocity is that the drag force resisting motion is directly proportional to the square of its speed. At low speeds the drag is much less than the gravitational force and so the object accelerates. As it speeds up the drag increases, until eventually it equals the weight. Drag also depends on the cross-sectional area. This is why things with a large surface area such as parachutes and feathers have a lower terminal velocity than small objects like bricks and cannon balls.
Mathematically, terminal velocity is described by the equation
V_t= \sqrt{\frac{2mg}{\rho A C_d }}
where
Vt is the terminal velocity,
m is the mass of the falling object,
g is gravitational acceleration,
Cd is the drag coefficient,
ρ is the density of the fluid the object is falling through, and
A is the object's cross-sectional area.
This equation is derived from the drag equation by setting drag equal to mg, the gravitational force on the object.
Note that the density increases with decreasing altitude, ca. 1% per 80 m (see barometric formula). Therefore, for every 160 m of falling, the "terminal" velocity decreases 1%. After reaching the local terminal velocity, while continuing the fall, speed decreases to change with the local terminal velocity
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The terminal velocity of an object falling towards the ground, in non-vacuum, is the speed at which the gravitational force pulling it downwards is equal and opposite to the atmospheric drag (also called air resistance) pushing it upwards. At this speed, the object ceases to accelerate downwards and falls at constant speed. An object moving downwards without power at greater than the terminal velocity (for example because it previously used power to descend, it fell from a thinner part of the atmosphere or it changed shape) will slow down until it reaches terminal velocity.
the formula would the be as follows......
V= sqrt[(2W) divided by (Cd)(p)(A)]
V= velocity
p= gas density
A= frontal area
Cd= drag coefficient
W= weight
...........christine@loved_0506
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First we must find the terminal velocity, to see if the thing
reaches it before it reaches the ground. We set the aerodynamic drag
equal to the force of gravity:
F_drag = 0.5 c p v^2 A
c = coefficient of drag for a thin airfoil = 0.05 (or so)
p = density of air = 1 g/L
v = terminal velocity of penny = ?
A = frontal area of falling (edge-on, we assume) penny,
about 18 mm x 1 mm
F_gravity = m g
m = mass of penny = 2 g (a guess)
g = 9.8 m/s^2
v = Sqrt[(m g)/(0.5 c p A)] = 200 m/s ( >1 sig fig not justified )
Penny would reach this velocity in 200 m/s / 9.8 m/s^2 = 20 seconds,
and it would travel (from rest) 0.5 9.8 20^2 = 2,000 m in that time.
The Empire State isn't 2 km high, so we conclude the penny will
not reach terminal velocity.
Hence the velocity is found roughly by using
h = 0.5 g t^2 = height of building = 391 m to observation deck
v = g t
Gives v = g Sqrt[h/(0.5 g)] = 88 m/s = 200 MPH. Since drag will be
strong already at this speed, this must be an upper estimate and we
expect the actual velocity to be lower.
Well, you really need to ignore air resistance here. In reality, a penny
falling sich a distance through the air will begin to tumble, making its
path chaotic. When that happens, you really can't know in advance exactly
how fast it will be falling when it lands.
The easy way to calculate this is to first calculate the potential energy
of the penny at the top of the building (relative to its potential energy
at the sidewalk), then assume that when it reaches the sidewalk, all its
former potential energy is converted to kinetic energy. If the height if
the building is H and the mass of the penny is M, the potential energy of
the penny at the beginning is MgH, where g is the acceleration due to
gravity, 9.8 m/s^2 (or, more usefully for this problem, 9.8 N/kg). At the
end of the fall, the penny's kinetic energy is MgH = 0.5 M v^2, where v is
the magnitude of the velocity. You can solve this equation for v:
MgH = (0.5) M v^2
gH = (0.5) v^2
2gH = v^2
v = (2gH)^(0.5)
In other words, the velocity will be the square root of the quantity 2gH.
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