A 70 kg and a 97 kg skydiver jump from an airplane at an altitude of 4150 m, bot

Question

A 70 kg and a 97 kg skydiver jump from an airplane at an altitude of 4150 m, both falling in the pike the density of air is 1.21 kg/m2 and the drag coefficient of a skydiver in a pike position is o.7.) If each skydiver has a frontal area of 0.14 m2, calculate their terminal velocities (in m/s). 70 kg skydiver 107 97 kg skydiver 121 m/s X m/s How long will it take (in s) for each skydiver to reach the ground (assuming the time to reach terminal velocity is small)? 70 kg skydiver 386 97 kg skydiver 34

Explanation / Answer

terminal velocity v = sqrt(2mg / AC)

v = sqrt(2*m*9.81m/s^2 / 1.21kg/m^3*0.14m^2*0.7)

v = sqrt(m * 165.46m^2/kg·s^2)

for 70 kg skydriver :

v = sqrt(m * 165.46m^2/kg·s^2)

v = sqrt(70 * 165.46m^2/kg·s^2)

v = 107.6 m/s

for 97 kg skydriver :

v = sqrt(m * 165.46m^2/kg·s^2)

v = sqrt(97 * 165.46m^2/kg·s^2)

v = 126.69 m/s

for 70 kg skydriver:-

t = d / v

t = 4159 m / 107.6 m/s

t = 38.65 s

for 92 kg skydriver:-

t = d / v

t = 4159 m / 126.69 m/s

t = 32.83 s

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