during a baseball game, a player's ball hit into the outfield is modeled by the equation h=-16t^2 + 65t. A bird flying across the field follows the path modeled by the equation h=8t+20. let t be the time since the ball is hit and h be the height. what do the intersection points of the equations represent?

The intersections of the equations represent the times when the bird and the ball are at the same height.

8t+20=-16t^2+65t

16t^2-57t+20=0, using the quadratic equation for expediency:

t=(57±√4529)/32 and since we know t>0

t≈3.88 seconds (at a approximate height of 51 feet)

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lisboa lisboa · Answer 2 · 2019-07-20T08:06:31+02:00

Answer:

Intersection points are

(0.39,0) and (3.16,0)

Step-by-step explanation:

The equation of the outfield is [tex]h=-16t^2 + 65t[/tex]

The path equation is [tex]h=8t+20[/tex]

To find intersection of two equations , we make them equal and solve for t

[tex]-16t^2+65t= 8t+20[/tex]

Subtract 8t from both sides

[tex]-16t^2+57t=20[/tex]

Subtract 20 on both sides

[tex]-16t^2+57t-20=0[/tex]

Divide whole equation by -1

[tex]16t^2-57t+20=0[/tex]

Apply quadratic formula to solve for t

[tex]t=\frac{-b+-\sqrt{b^2-4ac}}{2a}[/tex]

a= 16, b=-57 and c=20

[tex]t=\frac{57+-\sqrt{(-57)^2-4(16)(20)}}{2(16)}[/tex]

[tex]t=\frac{57+-\sqrt{1969}}{32}[/tex]

[tex]t=\frac{57+\sqrt{1969}}{32}=3.16[/tex]

[tex]t=\frac{57-\sqrt{1969}}{32}=0.39[/tex]