Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 2 - Motion Along a Straight Line - Problems - Page 38: 93b

Answer

The maximum height reached above point B is $~~1.0~m$

Work Step by Step

The stone travels a distance of $y = 3.00~m$ between point A and point B. The initial velocity at point A is $v$ and the final velocity at point B is $\frac{1}{2}v$ We can find $v$: $v_f^2 = v_0^2+2ay$ $(\frac{1}{2}v)^2 = v^2+2ay$ $\frac{3}{4}v^2 = -2ay$ $v^2 = -\frac{8ay}{3}$ $v = \sqrt{-\frac{8ay}{3}}$ $v = \sqrt{-\frac{(8)(-9.8~m/s^2)(3.00~m)}{3}}$ $v = 8.854~m/s$ $\frac{1}{2}v = 4.43~m/s$ In the next part of the question, let $v_0 = 4.43~m/s$ and let $v_f = 0$ We can find the maximum height reached above point B: $v_f^2 = v_0^2+2ay$ $y = \frac{v_f^2-v_0^2}{2a}$ $y = \frac{0-(4.43~m/s)^2}{(2)(-9.8~m/s^2)}$ $y = 1.0~m$ The maximum height reached above point B is $~~1.0~m$.
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.