# Air Resistance Formulas for Several Scenarios with Example

Friction between air and another object is known as air resistance. Let us examine how to determine an object’s air resistance when it is falling.

A falling object’s air resistance can be calculated by multiplying the air density times the drag coefficient times the area by two and then multiplying by velocity.

Gravity and air resistance are the two natural field forces that move everything on Earth. The air resistance formula for a sphere, the air resistance formula proof, the air resistance formula for free fall, and how to get average air resistance, will all be covered in more detail.

## How to calculate air resistance of a falling object?

The speed, area, and shape of the object passing through the air all affect air resistance. Let us check how to estimate a falling object’s air resistance.

To determine how much air resistance of a falling item will experience, use the formula, FD = 1/2 ρv2CDA. In this equation, FD stands for drag, ρ is fluid density, v for relative object speed to the fluid, CD for drag coefficient, and A for cross-sectional area.

Problem: An enormous passenger jet is travelling at a speed of 250.0 meter per second. A = 500 square meter of the wings of the aircraft are exposed to the wind. The drag coefficient is CD = 0.024. The density of the air ρ = 0.4500 kilogram per cubic meter at the plane’s height. How much air resistance is the passenger jet being subjected to?

Solution: Given data are,

A = 500 square meter

CD = 0.024

ρ = 0.4500 kilogram per cubic metre

Air resistance of a falling item given by,

FD = 1/2 ρv2CDA

FD =(0.4500 kg/m3 × 0.025 × 510.0 m2)/2 (250.0 m/s) 2

FD = (0.4500 kg/m3 × 0.025 × 510.0 m2)/2 (62500 m2/s2)

FD = 179296 kg .m/s2

## How to calculate air resistance in projectile motion?

The object or particle is referred to as a projectile, and its motion is referred to as projectile motion. Let us see how air resistance in projectile motion can be calculated.

Velocity, acceleration, and displacement must all be included when describing projectile motion in its whole, as describe below,

• Along the x and y axes, we must locate their constituent parts. Assume that all forces are insignificant besides gravity.
• The components of acceleration are then extremely straightforward once the positive direction is defined as upward, ay = -g = – 0.98 m/s2 (-32 ft/s2).
• Since gravity is vertical, ax = 0. ax = 0 indicates that vx = v0x, or that the initial and final velocities in the x direction are equal.
• With these constraints on acceleration and velocity, the kinematic equation x (t) = x0 + (vx) avgt for motion in a uniform gravitational field can be written through the equation v2y (t) = v2oy + 2ay (y – y0), which also includes the remaining kinematic equations for motion with acceleration with a constant acceleration.
• The kinematic equations for motion in a homogeneous gravitational field become kinematic equations with ay = -g, ax = 0.
• Horizontal motion, v0x= vx, x = x0 + vxt.
• Vertical motion, y = y0 + ½ (v0y + vy)t; vy = voy – gt; y = yo + voyt – ½ gt2, v2y = v2oy – 2g (y – yo).

Problem: A shell is launched into the air during a fireworks show at an angle of 75.00 above the horizontal with an initial speed of 70.0 m/s. The shell is timed so that the fuse will set off right when it is at its peak elevation above the earth.

• a. Make a calculation for the shell’s explosion height.
• b. How long does it take for the shell to launch and explode?
• c. What happens to the shell’s horizontal position when it explodes?
• d. From the launch site to the highest point, how far has the object moved overall?

Solution: (a) By “height,” we refer to the height above the starting point, or the altitude. When vy = 0, the highest point in any trajectory, known as the apex, is attained. We use the following equation to get y because we know the initial location, initial and final velocities, and initial position:

v2y = v2oy – 2g (y – y0)

The equation is made simpler by the fact that yo and vy are both zero.

0 = v2oy – 2gy.

By figuring out y, we get, y = v2oy/2g.

Now we need to figure out what the initial velocity’s y component, or v0y, is. It may be calculated using the formula v0y=v0sin θ, where v0 denotes an initial velocity of 70.0 m/s and θo=75°denotes an initial angle. Thus-

v0y=v0sin θ = (70.0 m/s) sin750 = 67.6 m/s and-

y = (67.6 m/s)2 / 2(9.80 m/s2)

y = 233 m.

The starting vertical velocity and maximum height are both positive because up is positive, while the acceleration brought on by gravity is negative. A projectile with an initial vertical component of velocity of 67.6 – m/s will reach a maximum height of 233 m. Also keep in mind that the maximum height only depends on the vertical component of the initial velocity (neglecting air resistance).

(b) There are various ways to determine when the projectile reaches its highest point, as in many physics problems. The simplest approach in this situation is to use vy=v0y -gt. This equation becomes vy= 0 at the apex

0 = v0y− gt

or,

t = voy/g = (67.6 m/s) / (9.80 m/s2)

t = 6.90 s.

Another way of finding the time is by using, y = yo + ½ (v0y + vy) t.

(c) Air resistance is little, hence axe and ay are both equal to zero. And as was previously mentioned, the horizontal velocity is constant. As shown by the equations x=x0+vxt, where x0 equals zero, the horizontal displacement is equal to the horizontal velocity multiplied by time. Thus,

x =vxt,

When the vx is the x component of the velocity, is given by

vx = v0cosθ = (70.0 m/s) cos75°=18.1 m/s.

Since both motions have the same time t, x is

x = (18.1 m/s) × 6.90 s=125 m.

Without air resistance, horizontal motion has a constant speed. The horizontal displacement observed here might be helpful in preventing audience trauma from falling pyrotechnics fragments. Air resistance plays a significant role in the shell explosion, and many fragments fall immediately below.

(d) Finding the size and direction of the displacement at the highest position is all that is required here as the displacement’s horizontal and vertical components have already been calculated:

s = 125 î + 233 ĵ; |ŝ|=√ (1252 + 2332) = 264 m; Φ = tan -1 (233/125) = 61.8°

## How to calculate air resistance at terminal velocity?

Air resistance is equivalent in magnitude to the weight of the falling object at terminal velocity. Let us examine the method for calculating air resistance at terminal velocity.

• Using Newton’s Second Law for a falling object as our starting point, we can determine air resistance at terminal velocity: Fg + Far = ma.
• To determine air resistance at a given speed the two types of air resistance are as follows: Far = – bv alternatively, Far = – bv2.
• To calculate air resistance at terminal velocity Newton’s law is used to determine air resistance at terminal velocity because the acceleration is zero, mg – bv = 0; mg – bv2 = 0.
• To determine air resistance at a given speed the answer to the velocity problem is vT = mg/b. An alternative is that v= √(mg/b).

If m represents mass in kilograms, g is the square of the gravitational acceleration, and b is an arbitrary quantity.

Problem: When dropped from rest, a 55 kilogram object experiences the air resistance force determined by Far = -15v2. Determine the object’s terminal velocity.

Solution: Utilize the formula vT = √ (mg/b) to determine the terminal velocity for a resistive force of the form Far = -bv2. Adding to the equation, we obtain,

vT = √(55) × (9.81)/15)

vT = 5.99 m/s

## How to calculate air resistance coefficient?

The drag coefficient varies as a square ratio of the object’s relative velocity. Let us examine the air resistance coefficient calculation method.

The air resistance coefficient is calculated using the equation c = Fair /v2. In the calculation, Fair is the force resistance and c is the force constant in this equation. Fluids, typically water in a sporting environment, are also subject to frictional force, which is not just restricted to air.

Fluid resistance, air resistance, and drag all refer to the same thing.

Problem: If an object travelling at 22 ms-1 encounter 50 N of air resistance, what is the force constant?

Solution: Given data are,

v = 22 ms-1

Fair = 50 N

The formula for air resistance coefficient is,

c = Fair /v2

Replace the specified values in the formula above. Then,

c = 50/ (22)2

c = 0.103

## How to calculate air resistance of parachute?

The weight pulses down on the cord as the parachute opens. Let us examine how to determine a parachute’s air resistance.

• To determine a parachute’s air resistance The equation for a parachute’s drag force, also known as its wind resistance force, is FD = 1/2 ρv2CDA. Where, FD is the drag force, r is the density of air, Cd is the drag coefficient, A is the area of the parachute and v is the velocity through the air.
• To determine a parachute’s air resistance with the square of the velocity, drag rises.
• To determine a parachute’s air resistance there is no net force acting on the rocket when drag is equal to weight. F = D – W = 0.
• Cd = 2 Fd / ρv2A = W to determine the parachute’s air resistance.
• And finally V = sqrt (2W/Cdρ A) is used to determine a parachute’s air resistance.

When two items are compared, the ones with a higher weight, a lower drag coefficient, a lower gas density, or a smaller area move at a higher speed.

## How to find air resistance with mass and acceleration?

The only force that affects humans at first is gravity, which propels them at a rate of -9.8 m/s2. Let us see how air resistance can be computed using mass and acceleration.

• For finding air resistance with mass and acceleration, we can use some algebra to get the object’s acceleration in terms of the net external force and the object’s mass (a = F/m).
• The net external force (F = W – D) is equal to the difference between the weight and drag forces. The acceleration of the object is then given by a = (W – D) / m.

Problem: A car has a mass of about 29 kg and is moving from Kolkata to Rajasthan at 50 meters per second and the track is loaded with iron and weighs 84 kg. Determine the drag force of the car.

Solution: Given data are,

Acceleration = 50 m/s2

Weight = 84 kg

Mass = 29 kg

We know that, a = (W – D) / m

50 = (84 – D)/ 29

1450 = 84 – D

-D = 1450 – 84

D = – 1366 N

## Air resistance graph

When air specks collide with an object’s front, it slows down. Let us check out this graph of air resistance.

By reducing the angle of release, air resistance’s impact on a projectile’s horizontal component of the trajectory can be minimized. Distance and speed, or velocity, are inversely proportional.

## How to calculate air resistance from speed?

The more air particles that impact the object, its overall resistance increase with surface area. Let us investigate how to determine air resistance based on speed.

The formula used to determine air resistance from speed is c = Fv2. The force of air resistance is represented by F in the technique, the force constant is represented by c, and the object’s velocity is represented by v. There is a linear relationship between air resistance and air density.

A quadric relationship is created between speed and air resistance. The area of an object’s leading edge that is travelling through the air determines how much air resistance it will experience. Air resistance increases as area increases.

Problem: If an object’s air resistance is 34 N and the force constant is 0.04, what is its speed?

Given data are, Fair = 34 N and c = 0.04

The formula for air resistance is,

Fair = cv2

v2 = 34/0.04

v2 = 850

v = 29.15 m/s.

## How to calculate force of air resistance?

The force of air resistance is measured in Newton (N). Let us examine how to determine the air resistance force.

Fair = – cv2 is the equation used to determine the force of air resistance. Fair is the force resistance and c is the force constant in this equation. The negative sign shows that the object is moving in the opposite direction from the direction of air resistance.

Problem: The force constant for a plane travelling at 50 ms-1 is 0.05. Determine the air resistance.

Solution: Given data are,

Velocity of air, v = 50

Force constant, c = 0.05

The force of air is given by,

F = – cv2

F = (-) 0.05 × 50 × 50

F = – 125 N.

## Air resistance formula for sphere

The relationship between the force of resistance acting on the body and air resistance is inverse. Let us look at the sphere’s air resistance formula.

The air resistance coefficient for sphere-shaped materials can be calculated using the following formula: Cd = 2 Fd / ρv2A, where for sphere-shaped materials-

• Cd = the air resistance coefficient,
• Fd is the Newton-based air resistance,
• A is the plan form area in square meters,
• ρ = the sphere’s density expressed in kilograms per cubic meter,
• And a substance’s viscosity expressed in meters per second is known as v.

Problem: Air density is 0.4500 kg/m3, and an aeroplane flying at altitude has a velocity of 250 m/s. 500 m2 of the airplane’s wings are exposed to the wind. The aeroplane is being impacted by 168750 N of air resistance. Do the drag coefficient calculation.

Solution: Given data, Air resistance for sphere-shaped materials, Fd = 168750 N

Density, ρ = 0.4500 kg/m3

Cross sectional are, A = 500 m2

Velocity, v = 250 m/s

We know that for sphere-shaped materials,

Cd = 2 Fd / ρv2A

Cd = 2 × 168750 / (0.4500 ×2502 × 500)

Cd = 0.025

## How to calculate average air resistance?

Air resistance is a kind of fluid friction that affects falling objects in the air. Let us see how to determine the average air resistance.

By multiplying the air density, drag coefficient, area, and velocity by two, one may calculate the average air resistance that a falling object will experience. Gravity causes objects to travel downward, in opposed to air friction, which acts in the opposite way and slows speed.

Air resistance rises as surface area grows for objects falling.

#### Conclusion

Air resistance is the force that an object experiences as it passes through the air, where if a person moves faster, the air resistance force grows. The dimensionless drag coefficient CD, which is calculated as CD = FD/1/2 ρAv2 where ρ is the fluid density (in this case, air). The object’s cross-sectional area is A = (1/4) ΠD2, and its speed is v.

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