ttregoning2006
Answered

Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

A football is punted from a height of 2.5 feet above the ground with an initial vertical velocity of 45 feet per second.

Write an equation to model the height h, in feet, of the ball t seconds after it has been punted.
The football is caught at 5.5 feet above the ground. How long was the football in the air?

Sagot :

rspill6

Answer:

1.451 seconds.  Hurry.

Step-by-step explanation:

The position(height, h) of the foot ball will be given by the equation:

h = -32.2t^2 + 45t + 2.5

The term -32.2t^2 consists of the acceleration caused by gravity at Earth's surface ( -32 ft/sec^2) and the term +45t accounts for the initial velocity.  The + is assigned since the ball is moving vertically from ground.  The negative sign in -32t^2 is due to the acceleration being towards ground.  The term 2.5 accounts for the fact that the initial height of the ball is 2.5 feet above ground.

We can either calculate or graph with this function to find the time the ball spends in the air:

Calculate:

h = -32.2t^2 + 45t + 2.5

We need a value of t that would cause h to be 5.5 feet, the level it was caught:

5.5 = -32.2t^2 + 45t + 2.5

-32.2t^2 + 45t + 2.5 = 5.5

-32.2t^2 + 45t -3.0  = 0

Solve for t using the quadratic formula.  I get a value of 1.33 seconds.

Graph:

See the attached graph.  It confirms the 1.33 second value

View image rspill6
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.