Let’s revisit our archery problem from earlier with a little more information.Andrew is an avid archer. He launches an arrow that takes a parabolic path. The equation of the height of the arrow with respect to time is y=−4.9x2+48x𝑦=−4.9𝑥2+48𝑥, where y𝑦 is the height of the arrow in meters above Andrew’s bow and x𝑥 is the time in seconds since Andrew shot the arrow. Find how long it takes the arrow to come back to a height even with his bow height.Let’s graph the equation by making a table of values. Remember it is helpful to start with finding the vertex so that we can know what values to use for x𝑥.y=−4.9x2+48x𝑦=−4.9𝑥2+48𝑥Find the x𝑥-value of the vertex by using the formula −b2a−𝑏2𝑎.−482(−4.9)≈4.9−482(−4.9)≈4.9Note: We used the ≈≈ symbol instead of the == sign because we are having to round this value. Remember we aren’t going to get as many “pretty” numbers with real world problems.)Now we’ll substitute that into our equation to find the y𝑦-value of our vertex.y=−4.9(𝑦=−4.9( )2+48()2+48( )=117.6)=117.6Our vertex is approximately (( ,, )).We’ll create our table of values centered around our vertex (but will round to 55 for that spot in our table).x𝑥y=−4.9x2+48x𝑦=−4.9𝑥2+48𝑥00y=−4.9(0)2+48(0)=0𝑦=−4.9(0)2+48(0)=011y=−4.9(1)2+48(1)=43.1𝑦=−4.9(1)2+48(1)=43.122y=−4.9(2)2+48(2)=76.4𝑦=−4.9(2)2+48(2)=76.433y=−4.9(3)2+48(3)=99.9𝑦=−4.9(3)2+48(3)=99.944y=−4.9(4)2+48(4)=113.6𝑦=−4.9(4)2+48(4)=113.655y=−4.9(5)2+48(5)=117.6𝑦=−4.9(5)2+48(5)=117.666y=−4.9(6)2+48(6)=111.6𝑦=−4.9(6)2+48(6)=111.677y=−4.9(7)2+48(7)=95.9𝑦=−4.9(7)2+48(7)=95.988y=−4.9(8)2+48(8)=70.4𝑦=−4.9(8)2+48(8)=70.499y=−4.9(9)2+48(9)=35.1𝑦=−4.9(9)2+48(9)=35.11010y=−4.9(10)2+48(10)=−10𝑦=−4.9(10)2+48(10)=−10Here’s the graph of the function.The roots of the function are approximately x=0𝑥=0 sec and x=9.8𝑥=9.8 sec. The first root tells us that the height of the arrow was 00 meters above his bow after seconds (right before Andrew first shot it) which makes sense. The second root says that it takes approximately seconds for the arrow to return to the height of the bow.Remember that we already calculated our vertex, so we don’t need to inspect our graph to find it, but we can use it to confirm it. We can interpret our vertex to mean that at approximately seconds, the arrow reached a maximum/minimum height of meters.CheckQuestion 7
Question
Let’s revisit our archery problem from earlier with a little more information.Andrew is an avid archer. He launches an arrow that takes a parabolic path. The equation of the height of the arrow with respect to time is y=−4.9x2+48x𝑦=−4.9𝑥2+48𝑥, where y𝑦 is the height of the arrow in meters above Andrew’s bow and x𝑥 is the time in seconds since Andrew shot the arrow. Find how long it takes the arrow to come back to a height even with his bow height.Let’s graph the equation by making a table of values. Remember it is helpful to start with finding the vertex so that we can know what values to use for x𝑥.y=−4.9x2+48x𝑦=−4.9𝑥2+48𝑥Find the x𝑥-value of the vertex by using the formula −b2a−𝑏2𝑎.−482(−4.9)≈4.9−482(−4.9)≈4.9Note: We used the ≈≈ symbol instead of the == sign because we are having to round this value. Remember we aren’t going to get as many “pretty” numbers with real world problems.)Now we’ll substitute that into our equation to find the y𝑦-value of our vertex.y=−4.9(𝑦=−4.9( )2+48()2+48( )=117.6)=117.6Our vertex is approximately (( ,, )).We’ll create our table of values centered around our vertex (but will round to 55 for that spot in our table).x𝑥y=−4.9x2+48x𝑦=−4.9𝑥2+48𝑥00y=−4.9(0)2+48(0)=0𝑦=−4.9(0)2+48(0)=011y=−4.9(1)2+48(1)=43.1𝑦=−4.9(1)2+48(1)=43.122y=−4.9(2)2+48(2)=76.4𝑦=−4.9(2)2+48(2)=76.433y=−4.9(3)2+48(3)=99.9𝑦=−4.9(3)2+48(3)=99.944y=−4.9(4)2+48(4)=113.6𝑦=−4.9(4)2+48(4)=113.655y=−4.9(5)2+48(5)=117.6𝑦=−4.9(5)2+48(5)=117.666y=−4.9(6)2+48(6)=111.6𝑦=−4.9(6)2+48(6)=111.677y=−4.9(7)2+48(7)=95.9𝑦=−4.9(7)2+48(7)=95.988y=−4.9(8)2+48(8)=70.4𝑦=−4.9(8)2+48(8)=70.499y=−4.9(9)2+48(9)=35.1𝑦=−4.9(9)2+48(9)=35.11010y=−4.9(10)2+48(10)=−10𝑦=−4.9(10)2+48(10)=−10Here’s the graph of the function.The roots of the function are approximately x=0𝑥=0 sec and x=9.8𝑥=9.8 sec. The first root tells us that the height of the arrow was 00 meters above his bow after seconds (right before Andrew first shot it) which makes sense. The second root says that it takes approximately seconds for the arrow to return to the height of the bow.Remember that we already calculated our vertex, so we don’t need to inspect our graph to find it, but we can use it to confirm it. We can interpret our vertex to mean that at approximately seconds, the arrow reached a maximum/minimum height of meters.CheckQuestion 7
Solution
The question seems to be incomplete. However, based on the information provided, it appears that you are asking for the time it takes for the arrow to return to the height of the bow.
From the given quadratic equation y = -4.9x^2 + 48x, we can find the roots of the equation to determine when the arrow is at the same height as the bow (i.e., when y = 0).
Setting y = 0, we get:
0 = -4.9x^2 + 48x
This can be factored to:
0 = x(-4.9x + 48)
Setting each factor equal to zero gives the solutions x = 0 and x = 48/4.9 ≈ 9.8.
So, the arrow is at the same height as the bow at the start (x = 0 seconds) and approximately 9.8 seconds after being shot. Therefore, it takes about 9.8 seconds for the arrow to return to the height of the bow.
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