Calculating Terminal Velocity for a skysurfer:
Skysurfer's attain a certain "terminal velocity" as they fall through the air, a speed at which one cannot fall any faster. The terminal velocity however differs from person to person and actually depends upon one's weight and the "shape" one assumes during free fall(The amount of air resistance incurred changes as person's shape or surface area changes). Terminal velocity is attained when one's weight equals the drag force imposed by air resistance.
ACCELERATION
a = F(net) = W - R m m
Where:
a = acceleration
W = weight
R = air resistance
m = mass
F(net) = net force acting on object
WEIGHT=DRAG
(at terminal velocity)
W = mg = D
Where:
W = Weight
m = mass
g = acceleration due to gravity
d = Drag
D = 1/2(Fl)(v)(v)(S)(cd)
Where:
D = Drag
Fl = Atmospheric Density
v = velocity
S = Cross sectional area
cd = Drag coefficient(an emperically derived dimensionless number, which relates to the shape of the object)
As shown by the formula above, the force due to air resistance which is also called drag, is directly proportional to density(Fl), cross sectional area(S), and drag coefficient(cd), while it is proportional to the square of the velocity. This essentially means that a change in the density, area or drag coefficient will cause an equal change in drag but a change in the velocity will have a much greater effect on the drag. If we say, double the velocity the drag will increase four times.
Skysurfer's try and maximize the drag force as much as possible so that they reach terminal velocity in the shortest possible time, thereby giving them the rush as well as a longer ride. They do this by maximizing the surface area exposed to the air pushing upward by stretching their bodies out into a "U" shape. Some skysurfer's will even deploy drag chutes as soon as they exit the plane. These are short drag lines that are attached to the skysurfer and hang out behind them, that increase the surface area that the air molecules act on thereby increasing the drag even more. When the drag force equals the force created by his weight, they cancel each other out as they are acting in opposite directions and therefore acceleration at that point is zero. He has reached his terminal velocity which is the fastest speed a body of his weight can travel at in free fall. He will continue to fall at this velocity for the rest of his jump.
Acceleration = 0 when Net External Force = 0
Calculations and Results:
We accordingly set the two equations above as equal to one another, plug in the values for the other variables and solve for a value of terminal velocity.
Weight = Drag
mg = 1/2(Fl)(v)(v)(S)(cd)
For our purposes we assume an average person of:
Mass(m): 150lbs
Acceleration due to Gravity(g):(32 ft/sec2)
Effective Area(S): Approx 9 sq.ft.
Drag Coefficient(cd): 0.7(fixed number)
Mass density(Fl):0.00236 slugs/(ft3) (standard value)
Upon solving this equation we find terminal velocity (v) to be 142 ft/sec or 98.8 mph. Correcting the density for a 6000 ft. average altitude, gives us a very realistic number, 115.8mph.