A sky diver of mass 70 Kg, and cross sectional area of 0.2 m2 is falling under gravity. Calculate the terminal velocity of this person. (Assume C= 0.88 kg/m3).

The terminal velocity of this person = 79.77 m/s

Explanation:

For a body falling through a fluid such as air, the forces acting on the diver include the force of gravity and the drag force.

And the net force is responsible for the acceleration of the diver.

Net force = mg - Drag force

But at terminal velocity, there's no acceleration and net force = 0

Hence, mg = Drag force

Drag force is given by CρAv²/2

where C = drag coefficient = 0.88

ρ = density of the fluid = density of air = 1.225 kg/m³

A = Cross sectional Area facing the fluid = 0.2 m²

v = velocity of the diver = ?

mg = 70 × 9.8 = 686 N

686 = (0.88×1.225×0.2×v²)/2

v² = (686×2)/(0.88×1.225×0.2)

v² = 6363.64

v = 79.77 m/s

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm