Physics Technology Update (4th Edition)

Published by Pearson

Chapter 2 - One-Dimensional Kinematics - Conceptual Questions - Page 48: 16

4.5 m/s.

Work Step by Step

Assume that the trampoline surface is at the origin (x=0) of the x axis. It is a motion with constant acceleration. The equation is $x=x_{0}+v_{0}t+\frac{1}{2}at^{2}$. In this case (freely falling object), $a = -g$. $x_{0}=0$. Now, the movement equation is $x=v_{0}t-\frac{1}{2}gt^{2}$. Make $x=0$: $0=v_{0}t-\frac{1}{2}gt^{2}$ $t(v_{0}-\frac{1}{2}gt)=0$ $t=0$ or $v_{0}-\frac{1}{2}gt=0$. Solving the second equation for t: $t=2\frac{v_{0}}{g}$ $t=0$ is the beginning of the movement, on the trampoline surface moving upward. But, $t=2\frac{v_{0}}{g}$ is the time when the person is back to the trampoline surface. The velocity as a function of time (motion with constant acceleration): $v=v_{0}+at$. Make $a=-g$ and $t=2\frac{v_{0}}{g}$ $v=v_{0}-g\times2\frac{v_{0}}{g}=v_{0}-2v_{0}=-v_{0}$. That is, the final velocity, when the person returns to the trampoline surface, is the negative of the initial velocity. So, the initial and final speed, the magnitude of the velocity, are the same: 4.5 m/s.

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