College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 4 - Problems - Page 149: 25


He left the floor with a speed of $5.05~m/s$

Work Step by Step

We can assume that the speed at a height of 1.3 meters is 0. We can find the initial speed $v_0$: $v_f^2 = v_0^2+2a\Delta y$ $v_0^2 = v_f^2-2a\Delta y$ $v_0 = \sqrt{v_f^2-2a\Delta y}$ $v_0 = \sqrt{0-(2)(-9.80~m/s^2)(1.3~m)}$ $v_0 = 5.05~m/s$ He left the floor with a speed of $5.05~m/s$
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