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May 11th, 2017, 04:24 PM  #1 
Senior Member Joined: Feb 2016 From: seattle Posts: 346 Thanks: 10  how to find When does Delbert's book pass him on its way down?
Hi I know it says to use a calculator but I am hoping and guessing it's not to hard to find this answer or help me find it. Thanks! 
May 11th, 2017, 05:41 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,518 Thanks: 1240  $h=16t^2+20t+300$ Time it reaches the highest point is at the vertex ... $t=\dfrac{b}{2a}=\dfrac{20}{2(16)}=\dfrac{5}{8} \, sec$ $h\left(\dfrac{5}{8}\right)=16\left(\dfrac{5}{8}\right)^2+20\left(\dfrac{5}{8} \right)+300=\dfrac{25}{4}+\dfrac{25}{2}+300 = 306.25 \, ft$ Passes the starting point on the way down ... $300=16t^2+20t+300$ $0=4t(4t5) \implies t=\dfrac{5}{4} \, sec$ 
May 15th, 2017, 01:01 PM  #3 
Senior Member Joined: Feb 2016 From: seattle Posts: 346 Thanks: 10 
Thanks as always skeeter!


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