4. 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 = 0.005x² + 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? Round your answer to the nearest hundredth of a meter.

Question

4.   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 = 0.005x² + 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? Round your answer to the nearest hundredth of a meter.

A.) 18.07 m

B.) 35.96 m

C.) 9.04 m

D.) 71.93 m

I have no clue as to what is is.

Please show your work.

Thank You

Damon · Answer

When is y zero? x = [ -.45 +/- sqrt(.2025+.13) ] / .01 = 100 [ -.45 +/- .577 ] = 12.66 m and x = -102.7 m sorry, I get none of the above but I get 12.7 meters

Dan · Answer

I think you may have misread the question. Check to see if .005x^2 is actually -.005x^2. Then follow the directions above. When you graph the flight of an object it should be a parabola that opens downwards. Think of the graph as a side view of the rock where the x-axis is the ground. The first x-intercept is the spot where it goes out of the hole and passes ground level. The second x-intercept is where it lands on the ground. Use the quadratic formula or a graphing calculator to find the zeros.