University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.4 - The Derivative as a Rate of Change - Exercises - Page 144: 11



Work Step by Step

$$s(t)=15t-\frac{1}{2}g_st^2$$ - Find $v(t)$: $$v(t)=\frac{ds}{dt}=\frac{d}{dt}\Big(15t-\frac{1}{2}g_st^2\Big)=15-g_st$$ The ball will reach its maximum height when its velocity decreases to $0$, which means $$v(t)=0$$ $$15-g_st=0$$ $$g_st=15$$ But we are already given that the ball reached its maximum height $20$ sec after being launched, so $t=20$: $$20g_s=15$$ $$g_s=\frac{3}{4}(m/sec^2)$$
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