Calculate the radius of parachute that will slow a 70 Kg person to a terminal velocity of 8 m/s. The person tries to avoid leg injury after touch down by crouching down by 60 cm then calculate the total force at the point of contact?

To calculate the radius of the parachute needed to slow a person to a terminal velocity of 8 m/s, we can use the drag force equation:

Fd = 0.5 * ρ * v^2 * Cd * A

Where:
Fd = Drag force
ρ = Air density
v = Velocity
Cd = Coefficient of drag
A = Surface area

First, let's calculate the drag force using the given values:
For a person of mass 70 kg, the weight can be calculated as follows:
Weight = mass * gravity
Weight = 70 kg * 9.8 m/s^2
Weight = 686 N

At terminal velocity, the drag force equals the weight:
Fd = Weight
0.5 * ρ * v^2 * Cd * A = Weight
0.5 * ρ * (8 m/s)^2 * Cd * A = 686 N

Next, we need to find the surface area A. Assuming a circular parachute shape, the surface area can be calculated as the area of a circle:
A = π * r^2

We already know the terminal velocity (v = 8 m/s) and weight (Weight = 686 N). However, we still need values for the air density (ρ) and coefficient of drag (Cd) to proceed with the calculations.

Let's assume:
- Air density (ρ) = 1.2 kg/m^3 (typical value near the Earth's surface)
- Coefficient of drag (Cd) = 1 (a rough approximation for a parachute shape)

Now we can solve for the radius (r) using the given values:

0.5 * (1.2 kg/m^3) * (8 m/s)^2 * 1 * π * r^2 = 686 N

Simplifying the equation:

0.5 * 1.2 * 64 * π * r^2 = 686 N

38.4 * π * r^2 = 686 N

r^2 = 686 N / (38.4 * π)
r^2 ≈ 5.65 m^2

Taking the square root of both sides:

r ≈ √(5.65 m^2)
r ≈ 2.38 m

Therefore, the radius of the parachute needed to slow the 70 kg person to a terminal velocity of 8 m/s is approximately 2.38 meters.

To calculate the total force at the point of contact after crouching down by 60 cm, we need to consider the impulse. Impulse is the change in momentum, and the force acting over a given time is equal to the change in momentum.

The change in momentum can be calculated as follows:
Change in momentum = mass * change in velocity

Given:
Mass = 70 kg
Change in velocity = 8 m/s (terminal velocity) - 0 m/s (crouching velocity)
Change in velocity = 8 m/s

Change in momentum = 70 kg * 8 m/s
Change in momentum = 560 kg·m/s

The force at the point of contact can be calculated by dividing the change in momentum by the change in time.

However, the change in time is not provided in the question. Therefore, we cannot calculate the total force at the point of contact without the change in time.

To calculate the radius of the parachute that will slow a person to a given terminal velocity, we need to start with the drag equation:

Fd = 0.5 * ρ * v^2 * Cd * A

Where:
- Fd is the drag force acting on the person,
- ρ (rho) is the air density,
- v is the velocity of the person relative to the air,
- Cd is the drag coefficient, and
- A is the cross-sectional area of the person.

First, let's calculate the drag force acting on the person:

Fd = m * g

Where:
- m is the mass of the person (70 kg),
- g is the acceleration due to gravity (9.8 m/s²).

Fd = 70 kg * 9.8 m/s² = 686 N

Now, let's rearrange the drag equation to solve for the area:

A = Fd / (0.5 * ρ * v^2 * Cd)

However, we don't have the values for air density (ρ) and the drag coefficient (Cd). So, we need to make some assumptions.

Assuming the air density is approximately 1.2 kg/m³, and the drag coefficient for a person falling in a spread-eagle pose is about 1.0, we can substitute these values into the equation.

A = 686 N / (0.5 * 1.2 kg/m³ * (8 m/s)^2 * 1.0)

A = 686 N / (0.5 * 1.2 kg/m³ * 64 m²/s²)

A ≈ 7.13 m²

Since the cross-sectional area (A) of a parachute can be approximated as the area of a circle with radius (r), we can rearrange the formula for the area of a circle:

A = π * r^2

To solve for the radius, we rearrange the equation:

r^2 = A / π

r^2 = 7.13 m² / π

r ≈ √(7.13 / π) ≈ 1.69 m

So, the radius of the parachute is approximately 1.69 meters.

Now, let's calculate the total force at the point of contact after the person crouches down by 60 cm (0.6 meters):

The total force can be calculated by adding the person's weight and the force experienced during the crouch.

Total force = Weight + Force during crouch

Weight = m * g = 70 kg * 9.8 m/s² ≈ 686 N

Force during crouch = m * a

Where:
- m is the mass of the person (70 kg),
- a is the acceleration experienced during the crouch.

Since the person is crouching down, we can consider the deceleration to be uniform, and we have two distances involved: the distance traveled while decelerating and the final position after crouching.

To calculate the acceleration (a), we use the equation:

a = (v^2 - u^2) / (2 * s)

Where:
- v is the final velocity (0 m/s after crouching),
- u is the initial velocity (8 m/s),
- s is the total distance traveled while decelerating (0.6 m).

a = (0^2 - 8^2) / (2 * 0.6 m) = -42.67 m/s²

Force during crouch = m * a = 70 kg * -42.67 m/s² ≈ -2,987 N

Total force = Weight + Force during crouch = 686 N + (-2,987 N)

Total force ≈ -2,301 N

The negative sign indicates that the force is acting in the opposite direction to the person's weight.

V_t = SQRT(W/CA)

W= 70KG X 9.81M/S^2 = 687 N (neglecting W of parachute)

V_t = 14/ms ; C = 0.88kg/m^3

A= W/(C x V_t^2) = 687N/(0.88kg/m^3 x 196m^2/s^2)= 3.98 m^2

A= pi x R^2
R= SQRT(A/PI) = 1.3