Question

A sky diver with a mass of 70 kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be FD =kV2, where k = 0.25 N*s^2 / m^2.
a. Determine the maximum speed of free fall for the sky diver and the speed reached after 100 m of fall.
b. Plot the speed of the sky diver as a function of time and as a function of distance fallen.

Answer 1

Answer:

The correct answer is

The maximum speed of free fall for the sky diver is 52.41 m/s

and the speed reached after 100 m is 37.44 m/s

Explanation:

From Force = mass × acceleration

Aerodynamic drag force = FD = k·V², where k = 0.25 N·s²/m²

The maximum speed, v, of free fall where m·a = 0 at terminal velocity

Mass × Gravity - Aerodynamic drag force = ma = 0

∴ 70 kg × 9.81 m/s² - 0.25 N·s²/m²× v² =0

Therefore v = 52.41 m/s

The speed after 100 m of free fall is given by

V = $\sqrt{\frac{m*g}{k} (1 - e^{(\frac{2k}{m} (y-y_{0} ))} }$  

Substituting we have, where y - y₀ = 100 then v = 37.44 m/s

b) To plot the relation we have

Speed to time relation as v or s(t) = $\sqrt{\frac{m*g}{k} (1 - e^{(-\frac{k}{m} t)} }$

Where S(t) is he speed at time t

and the relation between time and distance = v × t

Please see the attached graphs