Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 3 - Kinematics in Two Dimensions; Vectors - General Problems: 56


6.6 m/s

Work Step by Step

Let x= 0 and y = 0 where the pebble starts, and let upward be the positive y direction. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. The final vertical velocity at the window is 0, and the vertical displacement is 8.0 m. Find the initial y velocity from Eq. 2.11c, 12.5 m/s. Use equation 2.11a to find the time for the pebbles to hit the window. This is 1.28 s. Finally, analyze the horizontal motion that takes place at constant horizontal velocity. 8.5 m = (V$_x$)(1.28 s) Solve for V$_x$ = 6.6 m/s. The pebbles are moving horizontally at this speed when they hit the window.
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