Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 2 - Exercises and Problems - Page 31: 37


a)123 m b) v= 39.2 m/s, d=44 m c)v=9.8 m/s, d = 118 m d)v=20 m/s, d = 103 m (Note, speed is a scalar and cannot be negative).

Work Step by Step

We know that: $ V_f^2 = V_0^2 +2ad$ Since the final velocity is 0, we know: $V_0^2+2ad=0$ We substitute g, the gravitational constant, for a and simplify: $2gd=V_0^2$ $d = V_0^2/2g$ $d = 122.5 m$ b-d) We now find the speed and altitude at given points in time. We know that the speed will be given by: $v = v_0+at$ $v_y = v_{0y}-gt$ We then find the altitude, given the equation given above: $V_f^2-V_0^2=2ad$ Using the answer obtained using the previous speed equation as the final speed, we find: $ d = \frac{V_f^2-V_0^2}{-2g}$ We plug in the given values to solve for the answer.
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