Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 13 - Section 13.4 - Motion in Space: Velocity and Acceleration - 13.4 Exercise - Page 879: 30

Answer

$s=\dfrac{3}{4}H$

Work Step by Step

Use velocity-time equation. $v=u+at$ This implies $t=\frac{u}{g}$ [ v=0] ...(1) Use position-time equation. $s=ut+\dfrac{1}{2}gt^2$ From equation (1), we have $s=u(\frac{u}{g})+\dfrac{1}{2}g(\frac{u}{g})^2$ $s=\dfrac{u^2}{2g}=H$ ...(2) Also, $s=u \cdot (\dfrac{u}{2g})-\frac{g}{2}(\dfrac{u}{2g})^2$ This implies $s=\dfrac{3}{4}H$
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