# Falling humans

**Mathematics**

Falling Humans

A skydiver is descending with a constant velocity. Consider air resistance. A free-body diagram for this situation looks like this:

Newton’s Second Law

Newton’s second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables – the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

Newton’s second law of motion can be formally stated as follows:

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

F_{net}= m * aFinding Acceleration

the equation for gravitational force ,

F

_{grav}= m•gNewton’s Second Law of Motion

Newton’s second law of motion (F

_{net}= m•a) will be applied to analyze the motion of objects that are falling under the sole influence of gravity (free fall) and under the dual influence of gravity and air resistance.Free Fall and Air Resistance

An exploration of free fall, the motion of objects that encounter air resistance will also be analyzed.

Free Fall MotionFree fall is a special type of motion in which the only force acting upon an object is gravity. Objects that are said to be undergoingfree fall, are not encountering a significant force of air resistance; they are falling under the sole influence of gravity. Under such conditions, all objects will fall with the same rate of acceleration, regardless of their mass.The ratio (F

_{net}/m) is sometimes called thegravitational field strengthand is expressed as 9.8 N/kg (for a location upon Earth’s surface). The gravitational field strength is a property of the location within Earth’s gravitational field and not a property of an elephant nor a mouse. All objects placed upon Earth’s surface will experience this amount of force (9.8 N) upon every 1 kilogram of mass within the object. Being a property of the location within Earth’s gravitational field and not a property of the free falling object itself, all objects on Earth’s surface will experience this amount of force per mass. As such, all objects free fall at the same rate regardless of their mass. Because the 9.8 N/kg gravitational field at Earth’s surface causes a 9.8 m/s/s acceleration of any object placed there, we often call this ratio the acceleration of gravity.Free Fall and Air Resistance

An exploration of free fall, the motion of objects that encounter air resistance will also be analyzed.

Falling with Air ResistanceAs an object falls through air, it usually encounters some degree of air resistance. Air resistance is the result of collisions of the object’s leading surface with air molecules. The actual amount of air resistance encountered by the object is dependent upon a variety of factors. To keep the topic simple, it can be said that the two most common factors that have a direct affect upon the amount of air resistance are the speed of the object and the cross-sectional area of the object. Increased speeds result in an increased amount of air resistance. Increased cross-sectional areas result in an increased amount of air resistance.Why does an object that encounters air resistance eventually reach a terminal velocity? To answer this questions, Newton’s second law will be applied to the motion of a falling skydiver.

In the diagrams below, free-body diagrams showing the forces acting upon an 85-kg skydiver (equipment included) are shown. For each case, use the diagrams to determine the net force and acceleration of the skydiver at each instant in time.

The

Fand the_{net}= 833 N, downa = 9.8 m/s/s, downa = (F

_{net}/ m) = (833 N) / (85 kg) = 9.8 m/s/sThe

Fand the_{net}= 483 N, downa = 5.68 m/s/s, downa = (F

_{net}/ m) = (483 N) / (85 kg) = 5.68 m/s/sThe

Fand the_{net}= 133 N, downa = 1.56 m/s/s, downa = (F

_{net}/ m) = (133 N) / (85 kg) = 1.56 m/s/sThe

Fand the_{net}= 0 Na = 0 m/s/sa = (F

_{net }/ m) = (0 N) / (85 kg) = 0 m/s/s.As an object falls, it picks up speed. The increase in speed leads to an increase in the amount of air resistance. Eventually, the force of air resistance becomes large enough to balances the force of gravity. At this instant in time, the net force is 0 Newton; the object will stop accelerating. The object is said to have reached a

terminal velocity. The change in velocity terminates as a result of the balance of forces. The velocity at which this happens is called the terminal velocity.The amount of air resistance depends upon the speed of the object. A falling object will continue to accelerate to higher speeds until they encounter an amount of air resistance that is equal to their weight. A 150-kg skydiver weighs more (experiences a greater force of gravity), it will accelerate to higher speeds before reaching a terminal velocity. Thus, more massive objects fall faster than less massive objects because they are acted upon by a larger force of gravity; for this reason, they accelerate to higher speeds until the air resistance force equals the gravity force.

The amount of air resistance an object experiences depends on its speed, its cross-sectional area, its shape and the density of the air. Air densities vary with altitude, temperature and humidity. Nonetheless, 1.29 kg/m

^{3}is a very reasonable value. The shape of an object effects the drag coefficient (C). Values for various shapes can be found._{d}

Mass Density (kg/m^3) 1.29 Object Speed (m/s) 50.0 Drag Coefficient 0.50 X-sectional Area (m^2) 0.800

Case study:

By using the standard equation:

dy / dt = v

m ．dv / dt = – mg + 1/2 ． Cd ．A ．p ．v^2

where,

m is the mass of body,

Cd is the drag coefficient,

A is the cross-sectional of area,

p is the air density.

The free-fall time is only just over 210 seconds.

The speed of sound is a function of aire temperature.

Air density at 39km is 0.00446 kg/m^3.

An initial acceleration of about gm/s^2 will have applied and the falling body will have accelerated for at least 30s before the body reached 300m/s.

In 30s the body will have descended to about 35km, where the air density has doubled.

At terminal speed the body weight will equal the aerodynamic drag, given by the equation

Drag = 1/2 ． Cd ． p ． V^2 ． S

where,

Cd is the drag coefficient,

p is the air density,

V is the air speed,

S is the frontal area,

M is the Mach number.

Assume that,

the body weighs about 120kg (120g netwons),

a frontal area of about 1.4m^2,

a supersonic Cd of about 1.4,

By applying the equation to calculate the approximate air density which applied when the body was travelling at the quoted

maximum of 833.9 mph (373m/s).

This is about 0.0086kg/m^3, which is roughly the air density at 35km altitude.

Therefore, the body reached 373m/s (M=1.24) at about 35km and thereafter decelerated gradually, because of increasing air density with descreasing altitude.

Reference :

Mathematics TODAY , april 2013, page 90,92.

The physics classroom http://www.physicsclassroom.com/Class/newtlaws/U2l3e.cfm

**Rayleigh’s equcation**

By using dimensional analysis :

M = mass

L = length

T = time

Assuming that :

the most significant parmeters affecting the pressure distribution over a moving body immersed in a fluid are ,

D , its size dimensions in L

P , the air density , dimensions in ML^ -3

V , the flow speed , dimensions in LT ^-1

v , the kinematic viscosity , dimensions in L ^2 T ^-1

K = rp , the bulk elasticity of air. p = pressure , r = the ratio of specific heats for air.

K has dimensions of pressure , MLT ^-2

F , the aerodynmaic force , ( drag force , lift force )

The symbol for “directly proportional” is ∝

Rayleigh’s equation is ,

F ∝ PV ^2 D ^2 g1 ( v / VD ) g2 ( K / P V^2 )

If ,

a is the speed of sound in the fluid, dimensions in LT^-1

then ,

a^2 = rp / P = K / P

therefore ,

Reynolds number Re , Mach number M of the flow

F ∝ PV^2 D^2 f3 (Re) f4 ( M )

In aerodynamics and ballistics ,

Rayleigh’s equation is ,

F = 1/s PV^2 A f ( Re, M )

where A is area. f depends upon the particular force.

if F is the drag force , D , then f is the drag coefficient , CD ,

so that ,

D = 1/2 Pv^2 A CD ( Re , M , ∝ , B )

where the flow direction is specified by ∝ , B.

it may be assumed that ,

CD = CD ( M )

this assumption will be made when modelling a human free fall in the earth’s atmosphere.

Reynolds number has a significant effect at low flow velocities.

The jump simulation starts at zero , the velocity rapidly increases and the overall effect due to Reynolds number may be neglected.

**Ballistic models**

In a stable projectile ,

the wind vector is constant , the relative wind will vary because the velocity of the body is changing in magnitude and direction

throughout a ballistic trajectory. This means that the body has some incidence to the flow.

In statically unstable projectile will topple while in motion.

Gyroscopically stablisied projectiles matched to physical properties of the projectile.

Any latteral moment produces a gyroscopic torque such as spin , moment and torque axes are mutually orthogonal.

**In general case**

A simple trapezoidal rule can be applied except for the first interval, where the time span can be approximated as if

v^2 = 2 gx

where , g = 9.78 m/s^2 , x = distance from its starting point , v = downward velocity

there was a negligible resistance for a very small initial part of the motion.

This voids the apparent singularity in equation

dt = dx / v(x)

at the starting point, where the velocity is zero.

Reference :

Mathematics Today vol 49 No. 4 August 2013 , vol. 50 No.2 April 2014

## Recent Comments