Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Cumulative Review Exercises - Page 1128: 41


a) The time after which the ball strikes the ground is $6\text{ seconds}$. b) The ball reaches its maximum height of $\text{196 }\!\!~\!\!\text{ feet}$ in $\text{2}\text{.5 }\!\!~\!\!\text{ seconds}$.

Work Step by Step

(a) We know that when the ball hits the ground, the displacement covered will be zero. So, equate $\begin{align} & s\left( t \right)=0 \\ & -16{{t}^{2}}+80t+96=0 \end{align}$ $\begin{align} & -16{{t}^{2}}+80t+96=0 \\ & -{{t}^{2}}+5t+6=0 \\ & {{t}^{2}}-5t-6=0 \\ & {{t}^{2}}-6t+t-6=0 \end{align}$ $\begin{align} & t\left( t-6 \right)+1\left( t-6 \right)=0 \\ & \left( t-6 \right)\left( t+1 \right)=0 \end{align}$ $ t=6\text{ or }t=-1$ Since t cannot be negative so, $ t=6\text{ }$ Thus, the ball strikes the ground after $6$ seconds. (b) We have to differentiate $ s\left( t \right)=-16{{t}^{2}}+80t+96$ to get the value of maximum height as: $\begin{align} & s\left( t \right)=-16{{t}^{2}}+80t+96 \\ & \frac{ds\left( t \right)}{dt}=-32t+80 \\ \end{align}$ And for maximum height, evaluate $\frac{ds\left( t \right)}{dt}=0$ as: $\begin{align} & -32t+80=0 \\ & 32t=80 \\ & t=\frac{80}{32} \\ & t=2.5 \end{align}$ Therefore, at $ t=2.5$, the value of s is: $\begin{align} & s\left( t \right)=-16{{\left( 2.5 \right)}^{2}}+80\left( 2.5 \right)+96 \\ & =-16\left( 6.25 \right)+200+96 \\ & =-100+296 \\ & =196 \end{align}$ Thus, the ball reaches its maximum height of $196$ feet at $2.5$ seconds.
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