Question

A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = -0.06x^2 + 10.1x + 5, where x is the horizontal distance, in meters, from the starting point on the roof and y is the height, in meters, of the rocket above the ground. How far horizontally from its starting point will the rocket land?

Answer 1

Hello,

y=-0.06x²+10.1x+5
==>6x²-1010x-500=0
Δ=1010²+4*6*500=1032100=(1015,92...)²
x=(1010-1015.92...)/12 <0 exclude
or x=(1010+1015.92..)/12=168.826935... (m)

Answer 2

Answer:

The horizontal distance from the starting point when rocket lands is:

168.83 meters.

Step-by-step explanation:

We are given a function that models the height of the rocket when it is 'x' meters away from the starting point.

The function is:

$y=-0.06x^2+10.1x+5$

Now we are asked to find the horizontal distance from the starting point when the rocket will land.

i.e. we are asked to find 'x' ; when y=0

i.e.

Let y=0

$-0.06x^2+1.01x+5=0\\\\i.e.\\\\6x^2-1010x-500=0$

on solving the quadratic equation we obtain:

$x=-0.49360\ and\ x=\dfrac{125}{6}=168.83$

As the distance can't be negative.

Hence, the viable solution is:

x=168.83