Landon is standing in a hole that is 6.5 ft deep. He throws a rock and it goes up into the air, out of the hole, and then lands on the ground above. The path of the rock can be modeled by the equation y = -0.005x^2 + 0.45x - 6.5, where x is the horizontal distance of the rock, in meters, from Landon and y is the height, in meters, of the rock above the ground. How far horizontally from Landon will the rock land?
Please solve and explain. Thanks.
The answer for this question is D) 71.93
Why multiply it by 200?
My answer choices are:
18.07 ft
35.96 ft
9.04 ft
71.93 ft
To find the horizontal distance from Landon where the rock will land, we need to determine the value of x when y becomes equal to zero. This is because when the rock lands, its height above the ground will be zero.
Given the equation y = -0.005x^2 + 0.45x - 6.5, we can set y to zero and solve for x:
0 = -0.005x^2 + 0.45x - 6.5
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -0.005, b = 0.45, and c = -6.5. Substituting these values into the formula:
x = (-(0.45) ± √((0.45)^2 - 4(-0.005)(-6.5))) / (2(-0.005))
Simplifying further:
x = (-0.45 ± √(0.2025 - 0.13)) / (-0.01)
x = (-0.45 ± √0.0725) / (-0.01)
x = (-0.45 ± 0.269) / (-0.01)
Now, we have two possible values for x, obtained by both adding and subtracting 0.269:
x1 = (-0.45 + 0.269) / (-0.01)
x1 = -0.181 / (-0.01)
x1 = 18.1
x2 = (-0.45 - 0.269) / (-0.01)
x2 = -0.719 / (-0.01)
x2 = 71.9
Since we're dealing with the horizontal distance, the negative value (x1) doesn't make sense in this context. Therefore, the rock will land approximately 71.9 meters horizontally from Landon.
so it lands at y=0
0=-.005x^2+.45x-6.5
multiply all by 200
x^2-90x+1300=0
using the quadratic equation..
x= (90+-sqrt(8100-5200)/2=45+-1/2 sqrt(2800)=45+-sqrt(2700)=45+-51.96
we reject the negative solution
x=96.96 feet
check my math,I did it in my head.