A ball is launched upwards from a height of 5 feet. The ball's height as a function of time, in seconds, is modeled by the funct
ion h(t)=-16t^2+144t+5. What is the time interval in which the ball has a height greater than 325 feet?
1 answer:
The solution to the problem is:
4 < t < 5
Given that the height of the ball is given by the equation:
h(t) = 16t² + 144t + 5
h represents the height and t represent the time taken to achieve the height.
To determine the time interval for the height to be greater than 325 feet, we need to equate h(t) to 325, hence:
-16t² + 144t + 5 > 325
-16t² + 144t - 320 > 0
-t² + 9t - 20 > 0
-t² + 4t + 5t - 20 > 0
-t(t - 4) + 5(t - 4) > 0
(t - 4)(-t + 5) > 0
t - 4 > 0; -t + 5 > 0
t > 4, t < 5
Therefore the interval for the height greater than 325 ft is:
4 < t < 5
Find more at: brainly.com/question/17718371
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