A 5 -m-diameter parachute of a new design is to be used to transport a load from flight altitude to the ground with an average vertical speed of $3 \mathrm{m} / \mathrm{s}$. The total weight of the load and parachute is 200 N. Determine the approximate drag coefficient for the parachute.

Short Answer

Expert verified
The approximate drag coefficient for the parachute is 0.56.

Step by step solution

01

Determine the Surface Area of the Parachute

The surface area \( A \) of the parachute can be calculated using the formula for the area of a circle which is \( A = \pi \times (d/2)^2 \), where \( d \) is the diameter of the parachute. The diameter of the parachute is given as 5 m, so the surface area will be \( A = \pi \times (5/2)^2 =19.63 \, m^2 \).
02

Apply the Drag Force Equation

The parachute descends at a constant speed, meaning that the drag force is equal to the weight of the parachute. Hence, we can write the drag equation as such: \( 200 = 0.5 \times \rho \times 19.63 \times C_D \times (3)^2 \). Here, \( \rho \) is assumed to be the standard air density at sea level, \(1.225 m^{-3}\). Therefore, the equation becomes: \( 200 = 0.5 \times 1.225 \times 19.63 \times C_D \times 9 \).
03

Solve for the Drag Coefficient

Re-arranging the equation to find \( C_D \), we find that \( C_D = \frac{200}{0.5 \times 1.225 \times 19.63 \times 9} = 0.56 \).

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