Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 15 - Fluids - Problems and Conceptual Exercises - Page 536: 100

Answer

(a) $x=0.66 m$ (b) $y=0.206m$

Work Step by Step

(a) First, we find the initial velocity. $v_{\circ}=\sqrt{2gh}$ $\implies v_{\circ x}=\sqrt{2(9.8m/s^2)(0.39m-0.11m)}=2.34m/s$ We know that $y=y_{\circ}+v_{\circ}t+\frac{1}{2}a_yt^2$ We plug in the known values to obtain: $0=0.11m+(2.34m/s)sin 36^{\circ}t-\frac{1}{2}(9.8m/s^2)t^2$ This simplifies to: $t=0.346s$ Now $x=v_{\circ x}t$ We plug in the known values to obtain: $x=(2.34m/s)(\cos 36^{\circ})(0.346s)$ $x=0.66 m$ (b) We know that $v_y=v_{\circ}sin\theta-gt$ $\implies t=\frac{v_{\circ}sin\theta}{g}$ $\implies t=\frac{(2.34m/s)sin36^{\circ}}{9.8m/s^2}$ $\implies t=0.140s$ Now the maximum height can be determined as $y_{max}=y_{\circ}+v_{\circ y}t+\frac{1}{2}a_y t^2$ We plug in the known values to obtain: $y_{max}=0.11m+(2.34m/s)sin 36^{\circ}(0.140s)-\frac{1}{2}(9.8m/s^2)(0.140s)^2$ $\implies y_{max}=0.206m$
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