Question

The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the terminal velocity (in meters per second and kilometers per hour) of an 86.0 kg skydiver falling in a pike (headfirst) position with a surface area of 0.145 m2. (Assume that the density of air is 1.21 kg/m3 and the drag coefficient of a skydiver in a pike position is 0.7. For each answer, enter a number.)

Answer 1

Answer:

terminal velocity is;

v = 117.54 m/s

v = 423.144 km/hr

Explanation:

Given the data in the question;

we know that, the force on a body due to gravity is;

$F_g$ = mg

where m is mass and g is acceleration due to gravity

Force of drag is;

$F_d$ = $\frac{1}{2}$pCAv²

where p is the density of fluid, C is the drag coefficient, A is the area and v is the terminal velocity.

Terminal velocity is reach when the force of gravity is equal to the force of drag.

$F_g = F_d$

mg =  $\frac{1}{2}$pCAv²

we solve for v

v = √( 2mg / pCA )

so we substitute in our values

v = √( [2×(86 kg)×9.8 m/s² ] / [ 1.21 kg/m³ × 0.7 × 0.145 m²] )

v = √( 1685.6 / 0.122015 )

v = √( 13814.6949 )

v  = 117.54 m/s

v = ( 117.54 m/s × 3.6 ) = 423.144 km/hr

Therefore terminal velocity is;

v = 117.54 m/s

v = 423.144 km/hr