Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 47

Answer

(a) $h=\frac{[v_0~sin(\theta)]^2}{2g}$ (b) At the angle $30.0^{\circ}$: h = 14.4 m range = 99.8 m At the angle $45.0^{\circ}$: h = 28.8 m range = 115.2 m At the angle $60.0^{\circ}$: h = 43.2 m range = 99.8 m

Work Step by Step

(a) $2ah=v_y^2-v_{0y}^2$ $2(-g)h=v_y^2-v_{0y}^2$ $h=\frac{v_{0y}^2-0}{2g}$ $h=\frac{[v_0~sin(\theta)]^2-0}{2g}$ $h=\frac{[v_0~sin(\theta)]^2}{2g}$ (b) Let $\theta = 30.0^{\circ}$: $h=\frac{[v_0~sin(\theta)]^2}{2g}$ $h=\frac{[(33.6~m/s)~sin(30.0^{\circ})]^2}{(2)(9.80~m/s^2)}$ $h = 14.4~m$ $range = \frac{v_0^2~sin(2\theta)}{g}$ $range = \frac{(33.6~m/s)^2~sin(2\times 30.0)^{\circ}}{9.80~m/s^2}$ $range = 99.8~m$ Let $\theta = 45.0^{\circ}$: $h=\frac{[v_0~sin(\theta)]^2}{2g}$ $h=\frac{[(33.6~m/s)~sin(45.0^{\circ})]^2}{(2)(9.80~m/s^2)}$ $h = 28.8~m$ $range = \frac{v_0^2~sin(2\theta)}{g}$ $range = \frac{(33.6~m/s)^2~sin(2\times 45.0)^{\circ}}{9.80~m/s^2}$ $range = 115.2~m$ Let $\theta = 60.0^{\circ}$: $h=\frac{[v_0~sin(\theta)]^2}{2g}$ $h=\frac{[(33.6~m/s)~sin(60.0^{\circ})]^2}{(2)(9.80~m/s^2)}$ $h = 43.2~m$ $range = \frac{v_0^2~sin(2\theta)}{g}$ $range = \frac{(33.6~m/s)^2~sin(2\times 60.0)^{\circ}}{9.80~m/s^2}$ $range = 99.8~m$
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