Question

if a ball is thrown directly upward with a velocity of 80 ft/s, its height (in feet) after t seconds is given by y = 80t - 16t². what is the maximum height attained by the ball? 80 feet 100 feet 25 feet 176 feet 50 feet

Explanation:

Step1: Find the derivative of the height function

The height function is \(y = 80t-16t^{2}\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), the derivative \(y^\prime=\frac{dy}{dt}=80-32t\).

Step2: Find the critical points

Set \(y^\prime = 0\), so \(80-32t = 0\). Solving for \(t\):

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Step3: Find the maximum height

Substitute \(t = \frac{5}{2}\) into the height function \(y = 80t-16t^{2}\).

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Answer:

B. 100 feet