Answer:
The diameter of the parachute should be approximately 7.952 meters.
Step-by-step explanation:
To calculate the diameter of the parachute, we need to determine the area of the parachute required to achieve a terminal velocity of 5 m/s for an object with a mass of 100 kg. We will use the formula for terminal velocity which involves the drag force balancing the gravitational force.
\(/(The drag force/) \( F_d \) is given by:\[F_d = \frac{1}{2} C_d \rho A v^2\]\)
Where:
\(- \( C_d \) is the drag coefficient (1.3 for the parachute).\\- \( \rho \) is the density of air (1.216 kg/m^{3} ).\\- \( A \) is the area of the parachute.\\- \( v \) is the terminal velocity (5 m/s).\)
The gravitational force \(\( F_g \)\) acting on the object is:
\(\[F_g = mg\]\)
Where:
\(- \( m \) is the mass of the object (100 kg).\\- \( g \) is the acceleration due to gravity (9.8 m/s^{2} ).\)
At terminal velocity, the drag force equals the gravitational force:
\(\[\frac{1}{2} C_d \rho A v^2 = mg\]\)
Solving for the area A:
\(\[A = \frac{2mg}{C_d \rho v^2}\]\)
Substituting the given values:
\(\[A = \frac{2 \times 100 \, \text{kg} \times 9.8 \, \text{m/s}^2}{1.3 \times 1.216 \, \text{kg/m}^3 \times (5 \, \text{m/s})^2}\]\)
\(\[A = \frac{1960 \, \text{kg} \cdot \text{m/s}^2}{1.3 \times 1.216 \times 25 \, \text{kg/m}^3 \cdot \text{m}^2/\text{s}^2}\]\)
\(\[A = \frac{1960}{39.52}\]\[A \approx 49.6 \, \text{m}^2\]\)
The parachute is hemispherical, so the area A corresponds to the flat circular base area of a hemisphere. The area of a circle is given by:
\(\[A = \pi r^2\]\)
Solving for the radius r:
\(\[r^2 = \frac{A}{\pi}\]\)
\(\[r = \sqrt{\frac{49.6}{\pi}}\]\[r \approx \sqrt{\frac{49.6}{3.1416}}\]\[r \approx \sqrt{15.8}\]\[r \approx 3.976 \, \text{m}\]\)
The diameter D is twice the radius:
\(\[D = 2r\]\[D \approx 2 \times 3.976\]\[D \approx 7.952 \, \text{m}\]\)
Therefore, the diameter of the parachute should be approximately 7.952 meters.
