Respuesta :
When an object is thrown upwards, it reaches a maximum height and then comes back to its starting position. If we put f(t) = 0, we can get the time the ball was thrown and the time it returned.
Solving:
-16t² + 94t + 12 = 0
t = -0.125 and t = 6; negative time is not possible so the first solution is discarded.
The maximum height lies at the time exactly between the two times of zero displacement. Max height at
t = (-0.125 + 6)/2 = 2.9375 s ≈ 3
3 < t < 6
Therefore, the answer is the first option.
Solving:
-16t² + 94t + 12 = 0
t = -0.125 and t = 6; negative time is not possible so the first solution is discarded.
The maximum height lies at the time exactly between the two times of zero displacement. Max height at
t = (-0.125 + 6)/2 = 2.9375 s ≈ 3
3 < t < 6
Therefore, the answer is the first option.
A reasonable domain of the graph of the function when the basketball falls from its maximum height to the ground is A. 3 < t < 6.
How to depict the function?
It should be noted that an object comes to the starting position when thrown upwards. From the information given, the function is f(t)=−16t2+94t+12.
We'll then solve for the value of t. This will be t = -0.125 and 6. The maximum height will then be:
t = (-0.125 + 6)/2
t = 3
Therefore, a reasonable domain of the graph of the function when the basketball falls from its maximum height to the ground is 3 < t < 6.
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