Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 3 - Kinematics in Two Dimensions - Problems - Page 74: 19

Answer

$(a)\space 4.37\space s$ $(b)\space 93.6\space m$

Work Step by Step

(a) Here we use equation 3.5b $S=ut+\frac{1}{2}at^{2}$ in the vertical direction to find the traveled time of the golf ball. $\uparrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation. $0=30.3\space m/s\times sin45t^{\circ}+\frac{1}{2}(-9.8\space m/s^{2})t^{2}$ $0=21.43\space m/s\space t-4.9\space m/s^{2}t^{2}$ $0=t(21.43-4.9\space s^{-1}t)$ Solving this quadratic equation provides two solutions, $t=0$ or $0=21.43-4.9\space s^{-1}t=>t=4.37\space s$ The first solution represents the situation when the golf ball just begins its flight, so we neglect this one. So, t = 4.37 s (b) Let's apply equation 3.5a $S=ut+\frac{1}{2}at^{2}$ in the horizontal direction to find the range of the golf ball. $\rightarrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation. $R=30.3\space m/s\times cos45^{\circ}\space \times 4.37\space s+0$ $R=93.6\space m$ Therefore range of the golfball = 93.6 m
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