Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.1 Exercises - Page 252: 56

Answer

7.102 meters

Work Step by Step

At maximum height, v=0 $v(t) = \int -9.8 dt = -9.8t+v(0) = 10-9.8t $ At max height, $v(t_0)= -9.8t_0+10 =0$ or $t_0 =10/9.8 =1.02$ At this time, the height can be calculated by $s(t)=\int (10-9.8t) dt = 10t - 4.9 t^2 + s(0) $ $s(t) = 10t - 4.9 t^2+ 2$ So, $s(t_0) = 10\times1.02 - 4.9 \times1.02^2+ 2 =\boxed{7.102}$
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