Physics Technology Update (4th Edition)

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

Chapter 4 - Two-Dimensional Kinematics - Problems and Conceptual Exercises - Page 109: 90

Answer

(a) $5.62m$ (b) $4.98m$ (c) $5.20m$

Work Step by Step

(a) We can find the required distance as follows: $t=\frac{x}{vcos\theta}$ $\implies t=\frac{3.05m}{(10.1m/s)cos75^{\circ}}$ $t=1.16s$ Now $y=vsin\theta-\frac{1}{2}gt^2$ We plug in the known values to obtain: $y=(10.1m/s)sin(75^{\circ})(1.16)-\frac{1}{2}(9.8m/s^2)(1.16s)^2$ $y=4.72m$ $d=\sqrt{x^2+y^2}$ $\implies d=\sqrt{(3.05)^2+(4.72)^2}$ $d=5.62m$ (b) We can find the required distance as $t=\frac{x}{vcos\theta}$ $t=\frac{4.75}{10.1cos(75^{\circ})}$ $t=1.18s$ Now $d=\sqrt{x^2+y^2}$ $d=\sqrt{(4.75)^2+(1.5)^2}$ $d=4.98m$ (c) We can find the distance just before the projectile lands as $R=\frac{v^2sin2\theta }{g}$ We plug in the known values to obtain: $R=\frac{(10.1)^2sin (150^{\circ})}{9.8}$ $R=5.20m$
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