Answer:
domain: [0, (6+√39)/4] ≈ [0, 3.06125]
Step-by-step 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
