Abaseball player hits a ball towards the left field wall of a ball park that is 330 feet from the home plate. using the equation

did the player hit the ball over the wall (homerun)? how far did the ball land(when is y=0)? what is the maximum height of the baseball? what is the height of the ball the moment it is hit?

Step-by-step explanation:

a) The height of the ball at x = 330 is ...

  y = (-0.008·330 +3)330 +3 = 0.36·330 +3 = 118.80 +3 = 121.80

The ball is 121.8 ft high at the distance of the wall. If the wall is higher than that, the ball did not go over the wall. We expect the wall is not that high, so the ball clears the wall for a home run.

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b) When y=0, the value(s) of the distance x can be found using the quadratic formula:

  0 = -0.008x^2 +3x +3

  x = (-3 ±√(3² -4(-0.008)(3)))/(2·(-0.008))

  = (3 ±√9.96)/0.016 ≈ 376.0 . . . feet (positive solution)

The ball landed 376 ft from home plate.

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c) The maximum of the quadratic can be found where x is -b/(2a) = -3/(2·(-.008)) = 3000/16 = 187.5.

The value of y at that distance is ...

  y = (-0.008·187.5 +3)187.5 +3 = 284.25 . . . feet

The maximum height is 284.25 ft.

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d) The height of the baseball at home plate (where x=0) is

  y = -0.008·0 +3·0 +3 = 3 . . . feet

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step-by-step explanation:

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