Question

A 78 kg skydiver can be modeled as a rectangular "box" with dimensions 24 cm × 35 cm × 170 cm . If he falls feet first, his drag coefficient is 0.80.What is his terminal speed if he falls feet first? Use ? = 1.2 kg/^m3 for the density of air at room temperature.

Answer 1

Answer:

The terminal speed of his is 137.68 m/s.

Explanation:

Given that,

Mass of skydiver = 78 kg

Area of box$A =24\times35=840\ cm$

Drag coefficient = 0.80

Density of air $\rho= 1.2\times kg/m^3$

We need to calculate the terminal velocity

Using formula of drag force

$F_{d} = \dfrac{1}{2}\rho v^2Ac$

Where,

$\rho$ = density of air

A = area

C= coefficient of drag

Put the value into the formula

$78\times9.8=\dfrac{1}{2}\times1.2\times v^2\times24\times10^{-2}\times35\times10^{-2}\times0.80$

$v^2=\dfrac{2\times78\times9.8}{1.2\times24\times10^{-2}\times35\times10^{-2}\times0.80}$

$v=\sqrt{\dfrac{2\times78\times9.8}{1.2\times24\times10^{-2}\times35\times10^{-2}\times0.80}}$

$v=137.68\ m/s$

Hence, The terminal speed of his is 137.68 m/s.