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A ball is thrown vertically upward with an initial velocity of 96ft/second. The distance of the ball from the ground after t seconds is modeled by the equation below. For what time range is the ball MORE than 128 feet above the ground?

s(t) = 96t – 16t2

User Wyrmwood
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1 Answer

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Final answer:

To find when the ball is above 128 feet over the ground, solve the inequality 96t - 16t² > 128. This yields 2 solutions, t = 2 s and t = 4 s, indicating the ball is over 128 feet between 2 and 4 seconds after launch.

Step-by-step explanation:

To determine the time range when the ball is more than 128 feet above the ground, we need to solve the inequality s(t) > 128, where s(t) = 96t - 16t² represents the height of the ball at time t. Setting up the inequality, we get:

96t - 16t² > 128

To solve this, we can rearrange terms to form a quadratic equation:

16t² - 96t + 128 = 0

Factoring gives us:

(4t - 8)(4t - 16) = 0

The solutions to this equation are t = 2 seconds and t = 4 seconds. This means the ball is at 128 feet at both 2 and 4 seconds, implying it is more than 128 feet above the ground between these times.

Hence, the ball is more than 128 feet above the ground for the time range t > 2 s and t < 4 s.

User Karan Khanna
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