Answer
Please see the work below.
Work Step by Step
As we know that
$v_{\circ}=\sqrt{2gh}$
The maximum range is given as
$R=\frac{v_{\circ}^2}{g}$
$R=\frac{(\sqrt{2gh})^2}{g}=2h$
Essential University Physics: Volume 1 (3rd Edition)
by Wolfson, Richard
Published by
Pearson
ISBN 10:
0321993721
ISBN 13:
978-0-32199-372-4
Chapter 3 - Exercises and Problems - Page 49: 77
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
