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?

Question

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.

Anonymous · Accepted Answer

The answer for this question is D) 71.93

bobpursley · Answer

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.

Jane · Answer

Why multiply it by 200? My answer choices are: 18.07 ft 35.96 ft 9.04 ft 71.93 ft