Question

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 function h(t)=-16t^2+144t+5. What is the time interval in which the ball has a height greater than 325 feet?​

Answer 1

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