A physics student stands at the top of a hill that has an elevation of 37 meters. He throws a rock and it goes up into the air and then falls back past him and lands on the ground below. the path of the rock can be modeled by the equation y = -0.2x^2 + 0.8x +37 where x is the horizontal distance, in meters, from the starting point on the top of the hill and y is the height, in meters, of the rock above the ground. How far horizontally from its starting point will the rock land?
a. 37.00
b. 67.43***
c. 27.43
d. 37.78
X = (-0.8 +- sqrt(0.64+2.96))/-0.04 =
-27.43, and 67.43
X = 67.43
-0.02x^2 Not -0.2x^2.
Your answer is correct.
To find out how far horizontally from its starting point the rock will land, we need to find the x-coordinate when y (height of the rock) is equal to zero.
Given equation: y = -0.2x^2 + 0.8x + 37
Setting y equal to zero and solving for x:
0 = -0.2x^2 + 0.8x + 37
We can solve this quadratic equation by factoring, completing the square, or by using the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values from the equation:
a = -0.2, b = 0.8, c = 37
x = (-0.8 ± √(0.8^2 - 4*(-0.2)*37)) / (2*(-0.2))
x = (-0.8 ± √(0.64 + 29.6)) / (-0.4)
x = (-0.8 ± √(30.24)) / (-0.4)
x = (-0.8 ± 5.5) / (-0.4)
Now let's evaluate both solutions:
x1 = (-0.8 + 5.5) / (-0.4) ≈ -17.43
x2 = (-0.8 - 5.5) / (-0.4) ≈ 22.43
Since we're looking for a positive value for x (distance cannot be negative), the rock will land approximately 22.43 meters horizontally from its starting point.
Therefore, the correct answer is c. 27.43.
To find how far horizontally from its starting point the rock will land, we need to determine the value of x when y equals zero. This is because when the rock lands, its height above the ground will be zero.
Given the equation for the path of the rock:
y = -0.2x^2 + 0.8x + 37
We can replace y with zero:
0 = -0.2x^2 + 0.8x + 37
To solve this equation, we can use the quadratic formula, which is:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
In this case, the equation is in the form of ax^2 + bx + c, so:
a = -0.2
b = 0.8
c = 37
Now we can substitute these values into the quadratic formula:
x = (-0.8 ± sqrt(0.8^2 - 4(-0.2)(37))) / (2(-0.2))
Simplifying further:
x = (-0.8 ± sqrt(0.64 + 29.6)) / (-0.4)
x = (-0.8 ± sqrt(30.24)) / (-0.4)
x = (-0.8 ± 5.5) / (-0.4)
Now we have two possible solutions:
x1 = (-0.8 + 5.5) / (-0.4) = 4.7 / (-0.4) = -11.75
x2 = (-0.8 - 5.5) / (-0.4) = -6.3 / (-0.4) = 15.75
Since we are looking for a positive value of x, the rock will land approximately 15.75 meters horizontally from its starting point.
Therefore, the answer is option c. 27.43.