A 52 kg and a 98 kg skydiver jump from an airplane at an altitude of 6000 m, bot

Question

A 52 kg and a 98 kg skydiver jump from an airplane at an altitude of 6000 m, both falling in the diving/headfirst position. Assume their surface area is 0.135 m2 and the drag coefficient is 0.70.

a)How long will it take for the first skydiver to reach the ground in seconds (assuming the time to reach terminal velocity is small and the density of the air is a constant 1.21 kg/m3)?

b)How long will it take for the second skydiver to reach the ground in seconds (assuming the time to reach terminal velocity is small)?

Explanation / Answer

m1 = 52 kg
m2 = 98 kg
A = 0.135 m^2

terminal velocity v = (2mg / AC)
v = (2*m*9.8m/s² / 1.21kg/m³*0.135m²*0.7) = (m * 171.4m²/kg·s²)

1) v = (52kg*171.4m²/kg·s²) = 94.4 m/s
and t = d / v = 6000m / 94.4m/s = 63.6 s

2) v = (98kg*171.4m²/kg·s²) = 129.6 m/s
and t = 6000m / 129.6m/s = 46.3 s

Hope this helps!

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