Answer:
26.4 m
Step-by-step explanation:
The path the ball follows is given by the function
[tex]y=-\dfrac{1}{4}x^2+\dfrac{33}{5}x[/tex]
Before it hits the ground means y will be zero.
[tex]0=-\dfrac{1}{4}x^2+\dfrac{33}{5}x\\\Rightarrow x(-\dfrac{1}{4}x+\dfrac{33}{5})=0\\\Rightarrow x=0, -\dfrac{1}{4}x+\dfrac{33}{5}=0\\\Rightarrow x=0, x=\dfrac{33}{5}\times4\\\Rightarrow x=0, 26.4[/tex]
There are two points where the vertical displacement is zero the point at which Nate tosses the ball and the point where it lands.
So, the ball travels 26.4 m horizontally before it hits the ground.
Solving the quadratic equation, it is found that the ball travels 26.4 feet horizontally before it hits the ground.
The height of the ball, after an horizontal distance of x feet, is given by:
[tex]y(x) = -\frac{1}{4}x^2 + \frac{33}{5}x[/tex]
Converting the fractions to decimal:
[tex]y(x) = -0.25x^2 + 6.6x[/tex]
It hits the ground when:
[tex]y(x) = 0[/tex]
Then:
[tex]-0.25x^2 + 6.6x = 0[/tex]
[tex]0.25x^2 - 6.6x = 0[/tex]
[tex]x(0.25x - 6.6) = 0[/tex]
[tex]x = 0[/tex], or:
[tex]0.25x - 6.6 = 0[/tex]
[tex]0.25x = 6.6[/tex]
[tex]x = \frac{6.6}{0.25}[/tex]
[tex]x = 26.4[/tex]
The ball travels 26.4 feet horizontally before it hits the ground.
A similar problem is given at https://brainly.com/question/10489198
