Answer:
Given equation of the parabola is,
[tex]h(x)=-\frac{1}{20}x^2+8x+6[/tex]
where h(x) is the height function and x denotes the horizontal distance.
To find the distance thrown does the ball hit the ground.
we get that,
[tex]\begin{gathered} h(x)=0 \\ -\frac{1}{20}x^2+8x+6=0 \end{gathered}[/tex]
To find the value of x, we get
[tex]-\frac{1}{20}x^2+8x+6=0[/tex]
In order to cancel the denominater multiply by -20, we get
[tex]x^2-160x-120=0[/tex]
Factorise the above equation, we get
[tex]x=\frac{160\pm\sqrt[]{(160)^2+4(120)}}{2}[/tex][tex]x=\frac{160\pm\sqrt[]{4(6400)+4(120)}}{2}[/tex][tex]x=\frac{160\pm2\sqrt[]{6400+120}}{2}[/tex][tex]x=80\pm\sqrt[]{6520}[/tex][tex]x=80\pm2\sqrt[]{1630}[/tex]
The distance between the initial and final point is,
[tex]d=\lvert80-2\sqrt[]{1630}-(80+2\sqrt[]{1630})\rvert[/tex][tex]d=4\sqrt[]{1630}[/tex]
Answer is:
[tex]d=4\sqrt[]{1630}\approx161.49[/tex]
