+28 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?
+2446 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. |
+2446 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. |
