Question

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?

Answer 1

1.451 seconds. Hurry.

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