Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 6 - Force and Motion-II - Problems: 36



Work Step by Step

We have that the terminal speed in spread eagle is 160km/h and that in nosedive it is 310km/h. Terminal speed is given by $v = sqrt(2F/CpA)$ where F is the gravitational force on the body, p is the fluid density, A is the cross sectional area and C is the drag coefficient. We are told to assume that C is constant in the two positions. We also know that F is constant because it is the same body and therefore the force of gravity will be equal. The p, or fluid density, is also the same because you are falling in air both times. Therefore, the only thing that changes is A. We can rewrite this as follows: $v =K* \sqrt(1/A)$ where K is all of the constants that are in the formula. Therefore, to find the ratio of the areas, we merely take the ratio of the velocities so that V1/V2 = $\sqrt(A2/A1)$. Since we are looking for the ratio of the slower position to the faster one, we will call position 2 the slower one. We then get that 310/160 = $sqrt(A2/A1)$ and solving for the ratio of A2/A1 we get 3.75
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