if a ball is thrown directly upward with a velocity of 80 ft/s, its height (in feet) after t seconds is…

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

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

Answer

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): [ \begin{align*} 80-32t&=0\ -32t&=- 80\ t&=\frac{-80}{-32}=\frac{5}{2} \end{align*} ]

Step3: Find the maximum height

Substitute (t = \frac{5}{2}) into the height function (y = 80t-16t^{2}). [ \begin{align*} y&=80\times\frac{5}{2}-16\times(\frac{5}{2})^{2}\ &=200 - 16\times\frac{25}{4}\ &=200-100\ &=100 \end{align*} ]

Answer:

B. 100 feet