Question

The air is less dense at higher elevations, so skydivers reach a high terminal speed. The highest recorded speed for a skydiver was achieved in a jump from a height of 39,000 m. At this elevation, the density of the air is only 4.3% of the surface density of air at 20∘C. Estimate the terminal speed of a skydiver at this elevation. Suppose the mass of the skydiver is 90 kg and the cross section area of the skydiver is 0.72 m2. You could consider the diver as a cylinder traveling sideways for this problem.

Answer 1

Answer:

the terminal velocity v_t= 202.96 m/s≅203 m/s

Explanation:

The expression for the terminal velocity

$v_t= \sqrt{\frac{2mg}{\rho AC_d} }$

here, C_d is the drag coefficient for the cylinder is 1.15

The surface density of the air at 20°C is

ρ_surface = 1.2041 kg/m^3

the density of air at an altitude of 39000 m

ρ= 4.3/100×39000 = 0.05177 kg/m^3

now substitute these values in equation above

we get

$v_t= \sqrt{\frac{2\times90\times9.81}{0.05177\times0.72\times1.15} }$

v_t= 202.96 m/s≅203 m/s

the terminal velocity v_t= 202.96 m/s≅203 m/s