University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 2 - Motion Along a Straight Line - Problems - Exercises: 2.87

Answer

An athlete spends a factor of 4.8 times more time above $\frac{y_{max}}{2}$. Than below $\frac{y_{max}}{2}$ for the duration of the jump.

Work Step by Step

Let $t_2$ be the time it takes to go from $\frac{y_{max}}{2}$ up to $y_{max}$. Note that $t_2$ is equal to the time it takes to fall freely from $y_{max}$ down to $\frac{y_{max}}{2}$. $\frac{y_{max}}{2} = \frac{1}{2}~g~t_2^2$ $t_2 = \sqrt{\frac{y_{max}}{g}}$ Let $t$ be the total time it takes to go from the floor up to $y_{max}$. Note that $t$ is equal to the time it takes to fall freely from $y_{max}$ down to the floor. $y_{max} = \frac{1}{2}~g~t^2$ $t = \sqrt{\frac{2~y_{max}}{g}}$ Let $t_1$ be the time it takes to go from the floor up to $\frac{y_{max}}{2}$. $t_1 = t - t_2 = \sqrt{\frac{2~y_{max}}{g}} - \sqrt{\frac{y_{max}}{g}}$ We will find the ratio of $t_2$ to $t_1$. $\frac{t_2}{t_1} = \frac{\sqrt{\frac{y_{max}}{g}}}{\sqrt{\frac{2~y_{max}}{g}} ~- ~\sqrt{\frac{y_{max}}{g}}}$ $\frac{t_2}{t_1} = \frac{1}{\sqrt{2} - 1} = 2.4$ We then multiple 2.4 by 2 to get 4.8, as this solution is only for the last period of the total jump. Therefore, an athlete spends a factor of 4.8 times more time above $\frac{y_{max}}{2}$ during the total jump compared to when he is below $\frac{y_{max}}{2}$ .
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