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 - Page 62: 2.49


37.6 m/s.

Work Step by Step

The rock has a constant downward acceleration. We know the displacement of zero and the time of 6.00 s. Use the kinematics formulas for constant acceleration to find the initial velocity. Use $y-y_o=v_{oy}t+0.5at^2$ $0= v_{oy}(6.00s)+0.5(-9.80m/s^2)(6.00s)^2$ $v_{oy}=29.4m/s$ Now that we have the initial velocity, find the final speed. $v_f^2=v_o^2+2a(\Delta x)$ $v_f^2=(29.4m/s)^2+2(-9.8m/s^2)(-28.0m)$ $v_f= 37.6 m/s$
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