To find the distance horizontally from Landon where the rock will land, we need to find the value of x when y = 0 (since the rock will land on the ground, which is at y = 0). So we set y = 0 in the equation:
0 = 0.005x² + 0.42x - 6.5
This is a quadratic equation in standard form (ax² + bx + c = 0), where a = 0.005, b = 0.42, and c = -6.5. We can use the quadratic formula to solve for x:
x = (-b ± √(b² - 4ac)) / 2a
Plugging in the values, we get:
x = (-0.42 ± √(0.42² - 4(0.005)(-6.5))) / 2(0.005)
Simplifying:
x = (-0.42 ± √(0.196 + 0.13)) / 0.01
x = (-0.42 ± 0.524) / 0.01
There are two possible solutions here:
x = (-0.42 + 0.524) / 0.01 = 10.23
x = (-0.42 - 0.524) / 0.01 = -104.6
The first solution makes sense (since we're looking for a positive distance from Landon), so the rock will land approximately 10.23 feet horizontally from Landon.
Therefore, the answer is c. 10.23 ft.
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 +
0.42x 6.5, where x is the horizontal distance of the rock, in feet, from Landon and y is the height, in feet, of
the rock above the ground. How far horizontally from Landon will the rock land?
a. 63.54 ft c. 10.23 ft
b. 20.46 ft d. 31.77 ftLandon 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 = 0.005x² +
0.42x- 6.5, where x is the horizontal distance of the rock, in feet, from Landon and y is the height, in feet, of
the rock above the ground. How far horizontally from Landon will the rock land?
a. 63.54 ft c. 10.23 ft
b. 20.46 ft d. 31.77 ft
1 answer
1 answer