Question

In a baseball game, an outfielder throws a ball to the second baseman. The path of the ball is modeled by the equation y=-1/900(x-375/2)^2+705/16, where y is the height of the ball in feet after the ball has traveled x feet horizontally. The second baseman catches the ball at the same height as the height at which the outfielder released it. How far was the second baseman from the outfielder at the time he caught the ball?

Answer 1

The second base 375 feet away from the outfielder.

Explanation:

The given equation is

`y=-(1)/(900)(x-(375)/(2))^2+(705)/(16)`

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.

`y=-(1)/(900)(0-(375)/(2))^2+(705)/(16)`

`y=-(1)/(900)((140625)/(4))+(705)/(16)`

`y=-(625)/(16)+(705)/(16)`

`y=(80)/(16)`

`y=5`

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.

`5=-(1)/(900)(x-(375)/(2))^2+(705)/(16)`

`-(1)/(900)\left(x-(375)/(2)\right)^2=-(625)/(16)`

`\left(x-(375)/(2)\right)^2=(140625)/(4)`

`x-(375)/(2)=\pm \sqrt{(140625)/(4)}`

`x=(375)/(2)\pm (375)/(2)`

`x=0,375`

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.