Question

Ralph hit a softball into the air with an initial upward velocity of 48 feet per

second. The height h in feet of the ball above the ground can be modeled by
h = -16t² +48t + 3, where t is the time in seconds after Ralph hit th
softball.
What is the realistic domain for this function?

Answer 1

domain: [0, (6+√39)/4] ≈ [0, 3.06125]

Explanation:

You want the realistic domain of a height function h = -16t² +48t + 3.

Domain

The given polynomial is defined for all values of t. Since it represents height above the ground, and after the ball is thrown, only function values in the first quadrant are of interest: h ≥ 0, t ≥ 0.

The height becomes zero at t = (6+√39)/4 ≈ 3.061 seconds after the ball is hit. Hence the realistic domain is [0, (6+√39)/4], the interval in which the ball is above ground.

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Additional comment

We can solve this using the "completing the square" method.

t^2 -3t = 3/16 . . . . . . divide by -16, add 3/16

(t -3/2)^2 = 3/16 +9/4 . . . . . add 9/4 to complete the square

t = 3/2 ± √(39/16) = (6 ± √39)/4 . . . . . . take the square root, add 3/2

Only the positive value of the solution is applicable to this problem. That value is ...

h = 0 at t = (6 +√39)/4 ≈ 3.061249