Answer:
The second base 375 feet away from the outfielder.
Step-by-step explanation:
The given equation is
[tex]y=-\frac{1}{900}(x-\frac{375}{2})^2+\frac{705}{16}[/tex]
Where y is the height of the ball in feet after the ball has traveled x feet horizontally.
Put x=0, to find the initial height of the bal.
[tex]y=-\frac{1}{900}(0-\frac{375}{2})^2+\frac{705}{16}[/tex]
[tex]y=-\frac{1}{900}(\frac{140625}{4})+\frac{705}{16}[/tex]
[tex]y=-\frac{625}{16}+\frac{705}{16}[/tex]
[tex]y=\frac{80}{16}[/tex]
[tex]y=5[/tex]
Therefore the initial height of the ball is 5 feet.
The second baseman catches the ball at the same height as the height at which the outfielder released it.
Put y=5 at find the another value of x, for which the height of the ball is 5.
[tex]5=-\frac{1}{900}(x-\frac{375}{2})^2+\frac{705}{16}[/tex]
[tex]-\frac{1}{900}\left(x-\frac{375}{2}\right)^2=-\frac{625}{16}[/tex]
[tex]\left(x-\frac{375}{2}\right)^2=\frac{140625}{4}[/tex]
[tex]x-\frac{375}{2}=\pm \sqrt{\frac{140625}{4}}[/tex]
[tex]x=\frac{375}{2}\pm \frac{375}{2}[/tex]
[tex]x=0,375[/tex]
Therefore the second value of x is 375 for which the height of ball is 5.
Therefore the second base 375 feet away from the outfielder.
