Q:

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 equals negative 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 the ground. How far horizontally from Landon will the rock land? Round your answer to the nearest hundredth of a meter.

Accepted Solution

A:
we are given[tex]y=-0.05x^2+4.5x-6.5[/tex]where x is the horizontal distance of the rock in meters, from Landony is the height, in meters, of the rock above the groundwe know that when rock lands vertical distance will become 0so, we set y=0and then we can solve for x[tex]y=-0.05x^2+4.5x-6.5=0[/tex]Multiply both sides by 100[tex]-0.05x^2\cdot \:100+4.5x\cdot \:100-6.5\cdot \:100=0\cdot \:100[/tex][tex]-5x^2+450x-650=0[/tex]now, we can use quadratic formula [tex]x=\frac{-450\pm \sqrt{450^2-4\left(-5\right)\left(-650\right)}}{2\left(-5\right)}[/tex][tex]x=45-\sqrt{1895},\:x=45+\sqrt{1895}[/tex][tex]x=1.4684,x=88.531[/tex]now, we can find horizontal distance so, horizontal distance is [tex]=88.531-1.4684[/tex][tex]=87.06m[/tex]..................Answer