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# Need help solving this word problem please

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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?

Nov 28, 2017

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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.
Nov 29, 2017

#1
+2337
+3

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