Question

A 75 kg skydiver can be modeled as a rectangular "box" with dimensions 20 cm * 40 cm * 180 cm. what is his terminal speed if he falls feet first? use 0.8 for the drag coefficient.

Answer 1

Terminal speed=136.7 m/s

the formula for the terminal speed is given by :

V= $\sqrt{\frac{2 mg}{\rho a C}}$

m= mass=75 kg

a= area=0.2 x 0.4=0.08 m²

ρ= density of air=1.23 kg/m³

C= drag constant =0.8

so V= $\sqrt{\frac{2 *75*9.8}{1.23*0.08*0.8}}$

v= 136.7 m/s

Answer 2

As the problem depicts, a 75kg skydiver as rectangular box of dimentions 20cm*40cm*180cm, the terminal velocity if he falls feet first will be 13.72m/sec

What is Terminal Velocity?

When the resistance of the medium through which the body is travelling prohibits further acceleration, a freely falling body finally reaches a constant speed which is called as terminal velocity.

The terminal velocity is given by the following formula,

$v=\sqrt{\frac{2mg}{\rho{a}{c}} }$.............(i)

here,

m= mass of the body

g= gravitational accelaration

ρ= density of air

a= Projected area

c=Drag Constant

The density of air is 1.23kg/m³

Projected area (a) = 20*40

=800cm²=0.08m²

Putting the values in equation (i) we get,

$v=\sqrt{\frac{2*75*9.81}{1.23*0.08*0.8 }$

$v=\sqrt{18692.8354}$

$v=136.72m/sec$

So, the terminal velocity, if he falls feet first, is 13.72m/sec

To know more about Terminal Velocity

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