Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 3 - Exercises and Problems: 58

Answer

Please see the work below.

Work Step by Step

(a) We know that The position of the particle in the x direction is $x=x_{\circ}+v_{\circ}t+(\frac{1}{2})a_xt^2$ $x=0+11t+(\frac{1}{2})(-1.2)t^2$ $x=11t-0.6t^2$ When the particle crosses the y-axis then x value is zero $0=11t-0.6t^2$ This simplifies to : $t=18s$ (b) The position of the particle in y direction is $y=y_{\circ}+v_{\circ}t+(\frac{1}{2})a_yt^2$ We plug in the known values to obtain: $y=0+145+(-\frac{1}{2})(0.26)t^2$ $y=14t+0.13t^2$ At $t=18s$ $y=14(18)+0.13(18)^2$ $y=290m$
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.