Answer:
From 10 to 20 meters.
Step-by-step explanation:
1) If the sparrow has a parabolic trajectory then it must be actually:
[tex]y=-5\left ( x+6 \right )^{2}+10[/tex]
2) If the eagle is at 20 meters high, then we write:
[tex]y=20[/tex]
Since the exact x coordinate was not given.
But since it wants to get the sparrow
3) If we expand the equation we have:
[tex]-5x^{2}-60x-170=0\\\\X_{v}=\frac{-b}{2a}=\frac{60}{-10} = -6\\Y_{v}=\frac{-\Delta}{4a} =\frac{-200}{-20} =10[/tex]
Since the maximum point is equal to 10. The distance where the sparrow is flying ranges from 10 to 20 meters to the eagles spot.
[tex]10\leq d \leq20 \:or \:[10,20][/tex]
4) Since the x coordinate was not given then we can neither precisely calculate the distance where A is nor where B is located.
