Question

A 87 kgkg skydiver can be modeled as a rectangular "box" with dimensions 18 cmcm ×× 47 cmcm ×× 180 cmcm . If he falls feet first, his drag coefficient is 0.80. What is his terminal speed if he falls feet first? Use rho = 1.2 kg/m3 for the density of air at room temperature.

Answer 1

Answer:

Therefore the terminal velocity = 1.45 m/s

Explanation:

Terminal velocity: Terminal velocity is the highest velocity of an object when it falls from rest trough a media.

$V_t=\sqrt{\frac{2w}{c_d\rho A}}$

$V_t$= terminal velocity

w = weight of the object = mg

$c_d$ = drag coefficient=0.80

A= frontal area

$\rho$ = media density = 1.2 kg/m³

m = mass = 8 kg

g= acceleration due to gravity = 9.8 m/s²

Front area = length× breadth

                 = (18×47)cm²

                 =846 cm²

Therefore the terminal velocity

$v_t= \sqrt{\frac{2\times 87\times 9.8}{0.80\times 846 \times 1.2}$

    =1.45 m/s

Therefore the terminal velocity = 1.45 m/s