Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 57: 129

Answer

(a) After $1sec$ while thrown upwards and after $1.5sec$ while falling downwards. (b) The ball never reaches $48feet$. (c) The ball reaches its maximum height of $25feet$. (d) The ball would take $1.25sec$ to reach $25feet$. (e) It would hit the ground after $2.5sec$

Work Step by Step

We have a base formula: $h=-16t^2+v_{0}t$, but we are given initial speed of $v_{0}=40ft/s$, so we get a formula (which we will use later): $h=-16t^2+40t$ (a) Now, we have to measure the time it takes for a ball to reach $24ft$ (which is value of $h$). But we also have to note that a ball reaches this height $2$ times, once when its thrown upwards and second time when it falls down to the ground. $24=-16t^2+40t$ $-16t^2+40t-24=0$ //To simplify, we can divide the expression by -8 $2t^2-5t+3=0$ $t_{1} = \frac{5-\sqrt{1}}{4} =1 sec$ $t_{2}= \frac{5+\sqrt{1}}{4} = 1.5 sec$ (b) In this case $h=48$ : $48=-16t^2+40t$ $16t^2-40t+48=0$ //To simplify, divide by 8 $2t^2-5t+6=0$ $t=\frac{5±\sqrt{-23}}{4}$, It is undefined here as we cannot take square root of a negative number, which means that the ball has never reached $48 feet$. (c) Solution for a quadratic equation is always $2$ numbers, which means that the height reached in a specific amount of time gets the same number $2$ times (while thrown upwards and while falling downwards). However, we get only one height value when the object reaches its maximum point. which is when a Discriminant ($D$) is $0$: In case of: $-16t^2+40t-h=0$ $D=b^2-4ac$ $0=40^2-64h$ $64h=1600$ $h=25feet$ (d) As we found out in (c) maximum height is $25feet$, we will simply input this value to the quadratic formula we have and calculate time it takes to reach this height: $25=-16t^2+40t$ $16t^2-40t+25=0$ //We already know from (c) that $D$=0, so we will instantly calculate $t$ $t=\frac{40}{32}=\frac{5}{4}=1.25sec$ (e) The ball hits the ground when: $-16t^2+40t=0$ $t(40-16t)=0$ $t=0$ (Thrown point) $40-16t=0$ $t=\frac{40}{16}=\frac{5}{2}=2.5sec$
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