Question

I really need help with this........please only answer if you have the right answer. And please try to make it easy to understand cause I'm really bad at math!!

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

The model for this type of vertical motion is  h(t) = h(0) + v(0)t + (1/2)gt^2, where g is acceleration due to gravity, here -32 ft/(sec^2) because we're using the English system of measurement.

Inserting the given data results in:

5.5 ft = 2.5 ft + (45 ft/sec)t + (1/2)(-32 ft/(sec^2)t^2.

Note that this is a quadratic equation which can be solved in various ways.  I'll use the quadratic formula.

Subtracting 5.5 ft from both sides, we get  0 = -3 + 45t - 16t^2.  Then a = -16, b = 45 and c = -3.  The discriminant, b^2 - 4(a)(c), is 45^2 - 4(-16)(-3), or 1833.

The solutions are as follows:

      -45 plus or minus √1833        -45 plus or minus 42.8

t = --------------------------------------- = ------------------------------------

             2(-16)                                               -32

                                                   = 2.74 and a negative root.

Ignore the negative root, since we're measuring time, which is positive in this type of problem.

Thus, the ball will be in the air for 2.74 seconds.