Question

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

Answer 1

`v_(max)=52.38(m)/(s)`

`v_(100)=33.81`

Step-by-step explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

`\sum{F}=0=F_d-W`

`F_d=W`

`kv_(max)^2=m*g`

`v_(max)=\sqrt{(m*g)/(k)} =\sqrt{(70*9.8)/(0.25)}=52.38(m)/(s)`

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

`\sum{F}=ma=W-F_d`

`ma=W-F_d`

`ma=mg-kv_(100)^2`

`a=g-(kv_(100)^2)/(m)` (1)

consider the next equation of motion:

`a = ((v_(x)-v_0)^2)/(2x)`

If assuming initial velocity=0:

`a = (v_(100)^2)/(2x)` (2)

joining (1) and (2):

`(v_(100)^2)/(2x)=g-(kv_(100)^2)/(m)`

`(v_(100)^2)/(2x)+(kv_(100)^2)/(m)=g`

`v_(100)^2((1)/(2x)+(k)/(m))=g`

`v_(100)^2=(g)/(((1)/(2x)+(k)/(m)))`

`v_(100)=\sqrt{(g)/(((1)/(2x)+(k)/(m)))}` (3)

`v_(100)=\sqrt{(9.8)/(((1)/(2*100)+(0.25)/(70)))}`

`v_(100)=\sqrt{(9.8)/(((1)/(200)+(1)/(280)))}`

`v_(100)=\sqrt{(9.8)/(((3)/(350)))}`

`v_(100)=√(1,143.3)`

`v_(100)=33.81`

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

`v = v_0 +at`

as stated before, the initial velocity is 0:

`v =at` (4)

joining (1) and (4) and reducing you will get:

`(kt)/(m)v^2+v-gt=0`

solving for v:

`v=\frac{ \sqrt{1+(4gk)/(m)t^2}-1}{(2kt)/(m) }`

Plots: