Ireally need with only answer if you have the right answer. and try to make it easy to understand cause i'm really bad at 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?

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.

step-by-step explanation:

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