Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.4 Motion in Space: Velocity and Acceleration - 13.4 Exercises - Page 918: 24


a) $\approx 3591m$ b) $\approx 1631m$ c) 205 m/s or 204.8 m/s

Work Step by Step

(a) As we are given that $v_0=200 m/s$ and $\theta =60^\circ$ Then $v(t)=\lt 200 \cos 60^\circ, 200 \sin 60^\circ-9.8 t\gt$ or, $v(t)=\lt 100 ,100 \sqrt3-9.8 t\gt$ Also, $r(t)=\int v(t) dt$ or, $r(t)=\int [\lt 100 ,100 \sqrt3-9.8 t\gt]dt$ or, $r(t)=\lt 100t ,100 \sqrt3t-4.9t^2\gt$ Hence, $t=35.915$ Thus, Range: $R=35.915 * 100 \approx 3591 m$ (b) As we are given that $v_0=200 m/s$ and $\theta =60^\circ$ See part (a), which says that $t=35.915$ Hence, Maximum height$=100+1531 \approx 1631 m $ (c) As we are given that $v_0=200 m/s$ and $\theta =60^\circ$ Need to find Horizontal Speed . Thus, $s(t)=|v(t)|=\sqrt{(100)^2+(100 \sqrt3-9.8 (35.915))^2}$ or, $s(t)=205 m/s$ or 204.8 m/s
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