Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 357: 65

Answer

(a) $y(t) = 450-4.9~t^2$ (b) It takes $9.58~s$ for the stone to reach the ground. (c) The stone strikes the ground with a velocity of $93.9~m/s$ (downward). (d) It takes $9.09~s$ for the stone to reach the ground.

Work Step by Step

(a) $a(t) = \frac{dv}{dt} = -9.8$ $v(t) = -9.8~t+C_1$ Since $v(0) = 0$, then $C_1 = 0$ $v(t) = \frac{dy}{dt} = -9.8~t$ $y(t) = -4.9~t^2+C_2$ Since $y(0) = 450,$ then $C_2 = 450$ $y(t) = 450-4.9~t^2$ (b) We can find the time $t$ when $y(t) = 0$: $y(t) = 450-4.9~t^2 = 0$ $4.9~t^2 = 450$ $t^2 = \frac{450}{4.9}$ $t = \sqrt{\frac{450}{4.9}}$ $t = 9.58~s$ It takes $9.58~s$ for the stone to reach the ground. (c) When $t = 9.58~s,$ then $v = (-9.8)(9.58) = -93.9~m/s$ The stone strikes the ground with a velocity of $93.9~m/s$ (downward). (d) $v(t) = \frac{dy}{dt} = -5-9.8~t$ $y(t) = 450-5~t-4.9~t^2 = 0$ We can use the quadratic formula: $t = \frac{5 \pm \sqrt{(-5)^2-(4)(-4.9)(450)}}{2(-4.9)}$ $t = \frac{5 \pm \sqrt{25+8820}}{-9.8}$ $t = -10.11, 9.09$ Since $t$ is positive, the solution is $t = 9.09~s$ It takes $9.09~s$ for the stone to reach the ground.
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