Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 2 - Exercises and Problems: 39

Answer

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.