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: 23


(a) $\approx 3535m$ (b) $\approx 1531m$ (c) 200 m/s

Work Step by Step

(a) As we are given that $v_0=200 m/s$ and $\theta =60^\circ$ From question: $v(t)=\lt 200 \cos 60^\circ, 200 \sin 60^\circ-9.8 t\gt$ or, $v(t)=\lt 100 ,100 \sqrt3-9.8 t\gt$ As $r(t)=\int v(t) dt$ or, $r(t)=\int [\lt 100 ,100 \sqrt3-9.8 t\gt]dt$ or, $r(t)=\lt 100t ,100 \sqrt3t-4.9t^2\gt$ Need to find the range. $t=\dfrac{2 (200)(sin 60)}{g} \approx 3535m$ (b) As we are given that $v_0=200 m/s$ and $\theta =60^\circ$ Use part (a), we have $t=35.35$ Need to find the Maximum height$=100 \sqrt3(17.58)-4.9(17.68)^2$ or, $=\approx 1531 m$ (c) As we are given that $v_0=200 m/s$ and $\theta =60^\circ$ Speed is given by $s(t)=|v(t)|=\sqrt{(100)^2+(100 \sqrt3-9.8 t)^2}=200 m/s$
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