Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 13 - Vector Functions - 13.4 Exercises - Page 894: 24

Answer

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

Work Step by Step

(a) Given: $v_0=200 m/s$ and $\theta =60^\circ$ Since, we have $v(t)=\lt 200 \cos 60^\circ, 200 \sin 60^\circ-9.8 t\gt$ This implies that $v(t)=\lt 100 ,100 \sqrt3-9.8 t\gt$ As $r(t)=\int v(t) dt=\int [\lt 100 ,100 \sqrt3-9.8 t\gt]dt=\lt 100t ,100 \sqrt3t-4.9t^2\gt$ This gives, $t=35.915$ Range will be: $R=35.915 * 100 \approx 3591 m$ (b) From part (a), we have $t=35.915$ Thus, Maximum height$=100+1531 \approx 1631 m $ (c) Horizontal Speed is given by $s(t)=|v(t)|=\sqrt{(100)^2+(100 \sqrt3-9.8 (35.915))^2}=205 m/s$ or 204.8 m/s Hence, the required answers are: (a) $\approx 3591m$ (b) $\approx 1631m$ (c) 205 m/s or 204.8 m/s
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