University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 3 - Motion in Two or Three Dimensions - Problems - Exercises - Page 96: 3.52

Answer

The front of the boat should be a distance of 25.5 meters from the dock when the equipment is thrown.

Work Step by Step

We can find $v_{0y}$ when the equipment leaves the tower. $v_{0y} = v_0~sin(\theta) = (15.0~m/s)~sin(60.0^{\circ})$ $v_{0y} = 12.99~m/s$ We can find $v_y$ when the equipment lands on the deck. $v_y^2 = v_{0y}^2+2gy$ $v_y = \sqrt{v_{0y}^2+2gy}$ $v_y = \sqrt{(12.99~m/s)^2+(2)(-9.80~m/s^2)(-8.75~m)}$ $v_y = 18.45~m/s$ (toward the ground) We can find the time $t$ for the equipment's flight in the air. $t = \frac{v_y-v_{0y}}{g} = \frac{(-18.45~m/s)-(12.99~m/s)}{-9.80~m/s^2}$ $t = 3.208~s$ We can find the horizontal distance $x$ that the equipment travels. $x = v_x~t = (15.0~m/s)~cos(60.0^{\circ})(3.208~s)$ $x = 24.1~m$ We can find the horizontal distance $d$ that the boat travels. $d = v_x~t = (0.450~m/s)(3.208~s)$ $d = 1.44~m$ The total distance $D$ from the dock is x+d. $D = x+d = 24.1~m + 1.44~m = 25.5~m$ The front of the boat should be a distance of 25.5 meters from the dock when the equipment is thrown.
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