Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Explore thousands of questions and answers from a knowledgeable community of experts on our user-friendly platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

A ball is thrown into the air with an upward velocity of 44 ft/s. It’s height h in feet after t seconds is given by the function h=-16t^2 +44t+5. How long does it take the ball to reach its maximum height? What is the balls maximum height?

Sagot :

9514 1404 393

Answer:

  • 1.375 seconds
  • 35.25 feet

Step-by-step explanation:

A quadratic function y=ax^2 +bx +c is symmetric about the vertical line x=-b/(2a). That is the x-coordinate of the vertex (maximum or minimum).

Here, we have ...

  h = -16t^2 +44t +5

  t = -44/(2(-16)) = 11/8 = 1.375

It takes 1.375 seconds for the ball to reach its maximum height.

The height at that time is ...

  h = (-16(11/8) +44)(11/8) +5 = (-22 +44)(11/8) +5 = 121/4 +5 = 141/4 = 35.25

The maximum height is 35.25 feet.

View image sqdancefan
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.