Answer:
From 10 to 20 meters.
Step-by-step explanation:
1) If the sparrow has a parabolic trajectory then it must be actually:
![y=-5\left ( x+6 \right )^{2}+10](https://tex.z-dn.net/?f=y%3D-5%5Cleft%20%28%20x%2B6%20%5Cright%20%29%5E%7B2%7D%2B10)
2) If the eagle is at 20 meters high, then we write:
![y=20](https://tex.z-dn.net/?f=y%3D20)
Since the exact x coordinate was not given.
But since it wants to get the sparrow
3) If we expand the equation we have:
![-5x^{2}-60x-170=0\\\\X_{v}=\frac{-b}{2a}=\frac{60}{-10} = -6\\Y_{v}=\frac{-\Delta}{4a} =\frac{-200}{-20} =10](https://tex.z-dn.net/?f=-5x%5E%7B2%7D-60x-170%3D0%5C%5C%5C%5CX_%7Bv%7D%3D%5Cfrac%7B-b%7D%7B2a%7D%3D%5Cfrac%7B60%7D%7B-10%7D%20%3D%20-6%5C%5CY_%7Bv%7D%3D%5Cfrac%7B-%5CDelta%7D%7B4a%7D%20%3D%5Cfrac%7B-200%7D%7B-20%7D%20%3D10)
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.
![10\leq d \leq20 \:or \:[10,20]](https://tex.z-dn.net/?f=10%5Cleq%20d%20%5Cleq20%20%5C%3Aor%20%5C%3A%5B10%2C20%5D)
4) Since the x coordinate was not given then we can neither precisely calculate the distance where A is nor where B is located.