University Physics with Modern Physics (14th Edition)

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

Chapter 3 - Motion in Two or Three Dimensions - Problems - Discussion Questions - Page 92: Q3.9

Answer

See work.

Work Step by Step

The velocity vector at maximum height is horizontal, with magnitude $v_ocos \theta_o$. The vertical velocity component is zero, and the horizontal component is the same as at the time of launch. The speed at maximum height is $v_ocos \theta_o$. The vertical velocity component is zero, and the horizontal component is the same as at the time of launch. The acceleration vector at maximum height is vertically downward, with magnitude $9.80 m/s^2$. The vertical acceleration is constant.
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