A skydiver of mass 80.0 kg (including parachute) jumps off a plane and begins her descent. Throughout this problem use 9.80 m/s2 for the free-fall acceleration.
a. At some point during her free fall, the skydiver reaches her terminal speed. What is the magnitude of the drag force Fdrag due to air resistance that acts on the skydiver when she has reached terminal speed?
b. For an object falling through the air at a high-speed v, the drag force acting on it due to air resistance can be expressed as
F=Kv2
where the coefficient K depends on the shape and size of the falling object and on the density of air. For a human body, the numerical value for K is about 0.250 kg/m.
Using this value for K, what is the terminal speed vterminal of the skydiver?
c. When the skydiver descends to a certain height above the ground, she deploys her parachute to ensure a safe landing. Usually the parachute is deployed when the skydiver reaches an altitude of about 900 m (3000 ft). Immediately after deploying the parachute, does the skydiver have a nonzero acceleration?
d.When the parachute is fully open, the effective drag coefficient of the skydiver plus the parachute increases to 60.0 kg/m . What is the drag force Fdrag acting on the skydiver immediately after she has opened the parachute?
e. What is the terminal speed vterminal of the skydiver when the parachute is opened?

The skydiver in free fall has two forces acting on her: the force of gravity (acting downward) and the air resistance (acting upward). While the force of gravity is constant, the air resistance increases with speed: therefore, as the skydiver falls down, since her speed increases, the air resistance increases too until its value becomes equal to the magnitude of the force of gravity. From this point, the skydiver's acceleration becomes zero, and therefore she continues her fall at a constant velocity, called terminal speed.

Therefore, the terminal speed occurs when the magnitude of the air resistance is equal to the magnitude of the force of gravity:

[tex]F_{drag}=F_{grav}=mg[/tex]

where

m = 80.0 kg is the mass of the skydiver+parachute

[tex]g=9.80 m/s^2[/tex] is the acceleration due to gravity

Substituting, we find

[tex]F_{drag}=(80)(9.80)=784 N[/tex]

b)

The drag force for a body falling through the air is given by

[tex]F_{drag}=kv^2[/tex]

where

k is a coefficient that depends on the shape and size of the body, and on the air density

v is the speed of the body

As we said in part a), when a body reaches its terminal speed, the air resistance is equal to the force of gravity. Therefore:

[tex]F_{drag}=F_{grav}\\kv^2=mg[/tex]

Here we have

m = 80.0 kg

[tex]g=9.80 m/s^2[/tex]

k = 0.250 kg/m

Solving for v, we find the terminal speed:

[tex]v=\sqrt{\frac{mg}{k}}=\sqrt{\frac{(80)(9.8)}{0.250}}=56 m/s[/tex]

c)

Just before deploying the parachute, the skydiver has already reached the terminal speed: this means that at that moment the acceleration of the skydiver is zero, because the force of gravity (downward) is balanced by the air resistance (upward).

Then, the skydiver deploys the parachute. The parachute has a larger surface, therefore increasing the coefficient k of the air resistance formula. As a result, the magnitude of the air resistance becomes suddenly larger than that of the force of gravity.

As a result, there is now an upward net force acting on the skydiver, so a non-zero force: and therefore, according to Newton's second law of motion,

[tex]F=ma[/tex]

This means that now the skydiver has a non-zero acceleration (precisely, the direction of the acceleration is upward), and therefore she will start slowing down.

d)

The magnitude of the drag force acting on the skydiver is always given by

[tex]F_{drag}=kv^2[/tex]

where

k is the coefficient

v is the speed

Immediately after the skydiver opens the parachute, the speed is still her terminal speed, so

v = 56 m/s

Instaed, the coefficient has now became

k = 60.0 kg/m

Therefore, the new drag force acting on the skydiver and parachute is

[tex]F_{drag}=(60)(56)^2=1.88\cdot 10^5 N[/tex]

e)

After the skydiver opens the parachute, the drag force suddenly increases. As a result, the speed of the skydiver will decrease (because of the upward acceleration), because the drag force is now larger than the force of gravity.

However, as the speed decreases, the drag force decreases too, until at some point the drag force will become again equal to the force of gravity. When this occurs, the skydiver's acceleration will become zero again, and she will continue with a new constant velocity, a new terminal speed.

This will occur when the drag force is equal to the force of gravity, so:

[tex]F_{drag}=F_{grav}\\kv^2=mg[/tex]

where

k = 60.0 kg/m is the new coefficient

And solving for v, we find:

[tex]v=\sqrt{\frac{mg}{k}}=\sqrt{\frac{(80.0)(9.80)}{60}}=3.6 m/s[/tex]

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oeerivona oeerivona · Answer 2 · 2021-09-05T16:25:01+02:00

The drag force is the force that opposes relative motion in a fluid

The calculated values terminal velocity and drag force of the skydiver are;

a. The drag force is 784 Newtons

b. The terminal speed is 56 m/s

c. Yes

d. Immediately the parachute is opened, the drag force is 188,160 Newtons

e. When the parachute is opened, the terminal speed of the skydiver is approximately 3.615 m/s

The reason the above values are correct is as follows:

The known parameters are;

A skydiver jumps off a plane

The mass of the skydiver, m = 80.0 kg

The free-fall acceleration, g = 9.80 m/s²

a. Required:

The magnitude of the drag force when she reaches her terminal speed

Solution;

At the terminal speed, the acceleration is zero, and she is in uniform motion, from which according to Newton's first law of motion, the net force is zero therefore;

The downward force due to weight is equal to the upward force due to drag which gives;

The drag force, [tex]F_{drag}[/tex] = The weight of the skydiver, W = m × g

[tex]F_{drag}[/tex] = 80.0 kg × 9.80 m/s² = 784 N

The drag force, [tex]F_{drag}[/tex] = 784 Newtons

b. Required:

The value of the terminal speed

Solution:

The given equation for the drag force is F = K·v²

Where;

K = The shape and size of the object and density factor = 0.250 kg/m

v = The speed with which the object falls

At the terminal speed, [tex]v_{terminal}[/tex], we have;

F = [tex]F_{drag}[/tex] = K × [tex]\mathbf{v_{terminal}}[/tex]²

From part a., we have;

F = [tex]F_{drag}[/tex] = 784 N

Which gives;

∴ 784 = 0.250 × [tex]v_{terminal}[/tex]²

= √(784/0.250) = 56

The terminal speed, [tex]v_{terminal}[/tex] = 56 m/s

c. Required:

Whether or not the the skydiver has a non zero acceleration

Solution:

Yes, when the parachute is deployed, the drag force is increased and therefore, the velocity of the skydiver and the parachute combined is reduced , therefore, the acceleration changes becoming nonzero

d. Required:

The value of the drag force acting on the skydiver immediately after she has opened the parachute

Solution;

The given value of the drag coefficient, K = 60.0 kg/m

Immediately the parachute is opened while the skydiver is moving at the initial terminal speed of v = 56 m/s, we have;

Drag force, F = 60.0 kg/m × (56 m/s)² = 188.160 KN

The drag force, immediately the parachute is opened F = 188.160 KN

e. Required:

The value of new terminal speed with the parachute opened

Solution:

From the equation for the drag force, F = K·v², where k = 60.0 kg/m, we have;

At the terminal speed, F = [tex]F_{drag}[/tex] = 784 N

Therefore, at the terminal speed, [tex]v_{terminal}[/tex], we have;

[tex]F_{drag}[/tex] = K·[tex]\mathbf{v_{terminal}}[/tex]²

784 = 60.0 × [tex]v_{terminal}[/tex]²

[tex]\mathbf{v_{terminal}}[/tex] = √((784 N)/(60.0 kg/m)) ≈ 3.615 m/s

The terminal speed of the skydiver when the parachute is opened, [tex]v_{terminal}[/tex] ≈ 3.615 m/s

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