posted by Shannon .
The path of a large stone fired from a catapult can be modeled by y= -0.00545x^2+1.145x
where x is the distance the stone traveled (in yards) and y is the height of the stone (in yards).
1. find the distance the stone traveled.
2. find the maximum height of the stone.
AND (sorry there is more)
the path of a large arrow fired from a catapult can be modeled by y= -0.0044x^2+1.68x, where x is the distance the arrow traveled (in yards) and y is the height of the arrow (yards). Give the height of the castle wall, find the safest distance from the wall to launch an arrow over the wall.
1. the height of the wall is 120 yards
2. the height of the wall is 100 feet.
Okay thanks you so much! please try them! i am so bad at math.
Right now we are on quadratic formulas so i think your probably supposed to use that. Thanks!
i will do the first, you do the second
1. set y = 0
0 = -0.00545x^2 + 1.145x
0.00545x^2 - 1.145x = 0
x(0.00545x - 1.145) = 0
x=0 (at the start) or
x = 1.145/0.00545 = 210.1
2 max of y would be at the vertex of the parabola.
the x of the vertex is -b/2a = -1.145/(2(-0.00454)) = 105.05
then max y = -0.00545(105.05)^2 + 1.145(105.05)