Fundamentals of Physics Extended (10th Edition)

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

Chapter 4 - Motion in Two and Three Dimensions - Problems - Page 87: 53b

Answer

$v_0 = 31.9~m/s$

Work Step by Step

We can find $v_x$: $v_x = \frac{50.0~m}{4.00~s} = 12.5~m/s$ By symmetry, the ball takes $1.00~s$ to fall from the top of the wall to the point where the ball is caught. The total time of flight is $6.00~s$ The time to reach maximum height is $t_1 = 3.00~s$ We can find $v_{y0}$: $t_1 = \frac{v_{y0}}{g}$ $v_{y0} = g~t_1$ $v_{y0} = (9.8~m/s^2)(3.00~s)$ $v_{y0} = 29.4~m/s$ We can find the magnitude of the ball's initial velocity: $v_0 = \sqrt{(12.5~m/s)^2+(29.4~m/s)^2}$ $v_0 = 31.9~m/s$
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