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An arrow is shot with an initial upward velocity of 100 feet per second from a height of 5 feet above the ground. The equation h=−16t2+100t+5 models the height in feet t seconds after the arrow is shot. After the arrow passes its maximum height, it comes down and hits a target that was placed 20 feet above the ground. About how long after the arrow was shot does it hit its intended target?

Bunnie  Nov 28, 2017

Best Answer 

 #1
avatar+1493 
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The equation \(h=-16t^2+100t+5\) represents the height, in feet, (h) of the arrow after t seconds elapsed. The question is essentially asking after how seconds (t) will the arrow reach a height of 20 feet (h).

 

\(20=-16r^2+100t+5\) Subtract 20 from both sides.
\(-16t^2+100t-15=0\) Use the quadratic formula to find the solutions of this equation.
\(t=\frac{-100\pm\sqrt{100^2-4(-16)(-15)}}{2(-16)}\) Now, simplify from here.
\(t=\frac{-100\pm\sqrt{9040}}{-32}\) 16 is the largest perfect square factor of 9040, so we can simplify the radical further.
\(\sqrt{9040}=\sqrt{16*565}=\sqrt{16}\sqrt{565}=4\sqrt{565}\)  
\(t=\frac{-100\pm4\sqrt{565}}{-32}\) -100,4, and -32 all have a GCF of 4, so this answer can still be simplified further.
\(t=\frac{-25\pm\sqrt{565}}{-8}\)  
\(t_1=6.096\text{sec}\quad t_2=0.154\) Since the question specified that the arrow had to reach the height only after reaching its maximum altitude, then the only correct answer is 6.096 seconds.
   
   
TheXSquaredFactor  Nov 29, 2017
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1+0 Answers

 #1
avatar+1493 
+3
Best Answer

The equation \(h=-16t^2+100t+5\) represents the height, in feet, (h) of the arrow after t seconds elapsed. The question is essentially asking after how seconds (t) will the arrow reach a height of 20 feet (h).

 

\(20=-16r^2+100t+5\) Subtract 20 from both sides.
\(-16t^2+100t-15=0\) Use the quadratic formula to find the solutions of this equation.
\(t=\frac{-100\pm\sqrt{100^2-4(-16)(-15)}}{2(-16)}\) Now, simplify from here.
\(t=\frac{-100\pm\sqrt{9040}}{-32}\) 16 is the largest perfect square factor of 9040, so we can simplify the radical further.
\(\sqrt{9040}=\sqrt{16*565}=\sqrt{16}\sqrt{565}=4\sqrt{565}\)  
\(t=\frac{-100\pm4\sqrt{565}}{-32}\) -100,4, and -32 all have a GCF of 4, so this answer can still be simplified further.
\(t=\frac{-25\pm\sqrt{565}}{-8}\)  
\(t_1=6.096\text{sec}\quad t_2=0.154\) Since the question specified that the arrow had to reach the height only after reaching its maximum altitude, then the only correct answer is 6.096 seconds.
   
   
TheXSquaredFactor  Nov 29, 2017

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