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: 93a

Answer

$v = 8.85~m/s$

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 $v$: $v_f^2 = v_0^2+2ay$ $(\frac{1}{2}v)^2 = v^2+2ay$ $\frac{3}{4}v^2 = -2ay$ $v^2 = -\frac{8ay}{3}$ $v = \sqrt{-\frac{8ay}{3}}$ $v = \sqrt{-\frac{(8)(-9.8~m/s^2)(3.00~m)}{3}}$ $v = 8.85~m/s$
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