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Answer:
The horizontal distance the ball travels before it hits the ground is 22.[tex]\overline{857142}[/tex] feet
Step-by-step explanation:
The given parameter are;
The function modelling the path of the ball tossed by Nate y = -14·x² + 335·x
x = The horizontal distance the ball travels from Nate in feet
y = The height of the ball in feet
The line equation modelling the hill is y = 15·x
The point where the ball hits the ground is given by the point the graph of the equation for the path of the ball and the path of the model of the line of the hill meet as follows;
Ball path is y = -14·x² + 335·x
Hill path is y = 15·x
The point both paths meet and the ball hits the ground is -14·x² + 335·x = 15·x
Which gives;
-14·x² + 335·x - 15·x = 0
-14·x² + 320·x = 0
320·x - 14·x² = 0
x × (320 - 14·x) = 0
x = 0, or x = 320/14 = 22 6/7 = 22.[tex]\overline{857142}[/tex] feet
Therefore;
The horizontal distance the ball travels before it hits the ground = x = 22.[tex]\overline{857142}[/tex] feet.
The path of ball will be parabolic in nature .Parabolic path is defined as the angle of trajectory of a projectile.
Ball travel 22.86 feet horizontally before it hits the ground.
Since, The path of the ball is modeled by the function [tex]y = -14x^{2} + 335x[/tex]
where x is the ball’s horizontal distance from Nate in feet and y is the ball’s height in feet.
And hill is modeled by equation, [tex]y=15x[/tex]
To find out how much ball will travel horizontally before it hits ground , we have to find intersection of given two model equation.
[tex]y = -14x^{2} + 335x[/tex]
[tex]y=15x[/tex]
So, [tex]-14x^{2} + 335x=15x\\\\-14x^{2} + 335x - 15x = 0\\\\320x - 14x^{2} = 0\\\\x(320-14x)=0\\\\x=0, x=320/14=22.86[/tex]
Thus, Ball travel 22.86 feet horizontally before it hits the ground.
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