Consider a body falling through the air. The force due to air resistance is given by, F(air)=CAv^2, where C is the coefficient of air friction, A is the area facing the direction of the motion, and V is the velocity of the object with respect to the air. Take the value, C=0.88kgm^-3. If a body falls sufficiently far it will reach its terminalVelocityt, vt. ind this terminal velocity. in miles per hour, for a body of mass 70 kg and effective area of 0.2m^2.

To find the terminal velocity for a body falling through the air, we can start by using the equation for force due to air resistance: F(air) = CAv^2.

The force due to gravity acting on the body is given by F(gravity) = mg, where m is the mass of the body and g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth).

At terminal velocity, the force due to air resistance is equal in magnitude and opposite in direction to the force due to gravity. Thus, we can set F(air) equal to F(gravity).

CAv^2 = mg

Solving for v, the velocity, we get:

v^2 = mg / CA

v = √(mg / CA)

Given:
- C = 0.88 kg/m^3 (coefficient of air friction)
- A = 0.2 m^2 (effective area)
- m = 70 kg (mass of the body)
- g = 9.8 m/s^2 (acceleration due to gravity)

Plug in the values into the equation:

v = √((70 kg * 9.8 m/s^2) / (0.88 kg/m^3 * 0.2 m^2))

v = √(686 / (0.176))

v = √(3897.727)

v ≈ 62.43 m/s

To convert the velocity from meters per second to miles per hour, we can use the following conversion:

1 mile = 1609.34 meters
1 hour = 3600 seconds

v_mph = v * (3600 s/h) / (1609.34 m/mile)

v_mph = 62.43 m/s * (3600 s/h) / (1609.34 m/mile)

v_mph ≈ 139.73 mph

Therefore, the terminal velocity for the body with a mass of 70 kg and an effective area of 0.2 m^2 is approximately 139.73 miles per hour.

To find the terminal velocity of the body, we need to set the force of gravity equal to the force of air resistance:

mg = F(air)

Where:
m = mass of the body (70 kg)
g = acceleration due to gravity (9.8 m/s^2)

The force of air resistance is given by:

F(air) = CAv^2

Where:
C = coefficient of air friction (0.88 kg/m^-3)
A = effective area of the body facing the direction of motion (0.2 m^2)
v = velocity of the object with respect to the air (terminal velocity, v_t)

Now, we can substitute these values into the equation:

mg = CAv^2

Rearranging the equation to solve for v:

v^2 = mg / CA

Now, let's plug in the given values and solve for v:

v^2 = (70 kg)(9.8 m/s^2) / (0.88 kg/m^-3)(0.2 m^2)

v^2 = 682 m^2/s^2 / (0.88 kg/m^-3)(0.2 m^2)

v^2 = 682 m^2/s^2 / (0.176 kg)

v^2 = 3869.32 m^2/s^2

Finally, taking the square root of both sides, we get:

v = sqrt(3869.32) m/s

Since you want the terminal velocity in miles per hour, we need to convert the velocity from m/s to miles/hour. Conversion factors:

1 mile = 1609.34 meters
1 hour = 3600 seconds

So, 1 m/s = (1 mile / 1609.34 meters) * (3600 seconds / 1 hour)

Plugging in the values:

v_mph = (v m/s) * (1 mile / 1609.34 meters) * (3600 seconds / 1 hour)

v_mph = [sqrt(3869.32) * (1 mile / 1609.34)] * (3600)

v_mph = sqrt(3869.32) * (1 mile * 3600) / (1609.34)

Now, calculating the value:

v_mph = sqrt(3869.32) * (2246.398) / (1609.34)

v_mph ≈ 86.86 mph

Therefore, the terminal velocity of the body is approximately 86.86 miles per hour.