landon is standing in a hole that is 6.5 m 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=-0.05x^2+4.5x-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 ground. How far horizontally from Landon will the rock land? Round the answer to the nearest hundredth of a meter.
To find the horizontal distance from Landon where the rock will land, we need to determine the value of x in the equation y = -0.05x^2 + 4.5x - 6.5 when y = 0.
This is because when the rock lands on the ground, its height above ground is zero, so y = 0.
Now we can set up the equation:
0 = -0.05x^2 + 4.5x - 6.5
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
In our equation, a = -0.05, b = 4.5, and c = -6.5.
Substituting these values into the quadratic formula, we have:
x = (-4.5 ± √(4.5^2 - 4(-0.05)(-6.5)))/(2(-0.05))
Simplifying this equation further, we get:
x = (-4.5 ± √(20.25 - 1.3))/(0.1)
x = (-4.5 ± √(21.55))/(0.1)
Using a calculator or solving it manually, we find two possible solutions for x:
x ≈ 1.34 or x ≈ 12.66
Since the rock is thrown upwards and then lands on the ground, we are only interested in the positive value for x, which is approximately 12.66 meters.
Therefore, the rock will land approximately 12.66 meters horizontally from Landon.
it lands when the height is zero, so just solve
-0.05x^2+4.5x-6.5 = 0