Question

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.

Answer 1

we are given

$y=-0.05x^2+4.5x-6.5$

where

x is the horizontal distance of the rock in meters, from Landon

y is the height, in meters, of the rock above the ground

we know that when rock lands vertical distance will become 0

so, we set y=0

and then we can solve for x

$y=-0.05x^2+4.5x-6.5=0$

Multiply both sides by 100

$-0.05x^2\cdot \:100+4.5x\cdot \:100-6.5\cdot \:100=0\cdot \:100$

$-5x^2+450x-650=0$

now, we can use quadratic formula

$x=\frac{-450\pm \sqrt{450^2-4\left(-5\right)\left(-650\right)}}{2\left(-5\right)}$

$x=45-\sqrt{1895},\:x=45+\sqrt{1895}$

$x=1.4684,x=88.531$

now, we can find horizontal distance

so, horizontal distance is

$=88.531-1.4684$

$=87.06m$..................Answer