During halftime of a soccer game, a sling shot launched t-shirts at the crowd. a t-shirt is launched from a height of 3 feet with an initial upward velocity of 80 feet per second. the t-shirt is caught 43 feet above the field. how long will it take the t-shirt to reach its maximum height? what is the maximum height? what is the range of the function that models the height of the t-shirt over time?

Step-by-step explanation:

Ballistic motion is usually modeled by the quadratic function ...

  h(t) = -1/2g·t^2 +v0·t +h0

where g is the acceleration due to gravity, v0 is the initial upward velocity, and h0 is the initial height.

When units are in feet and seconds, the value of g is usually taken to be 32 ft/s^2, so for v0 = 80 ft/s and h0 = 3 ft, the equation is ...

  h(t) = -16t^2 +80t +3

The axis of symmetry of the graph of this equation is ...

  t = v0/g = (80 ft/s)/(32 ft/s^2)

  t = 2.5 s . . . . . the time to reach the maximum height

At that time, the maximum height is ...

  h(2.5) = (-16·2.5 +80)2.5 +3 = 103 . . . . feet

Then the range of the function extends from its minimum value, 3 at t=0, to its maximum value, 103 at t=2.5. Since the T-shirt is caught at h(t) = 43, the T-shirt never goes below 3 feet.

The range of the function is 3 ft to 103 ft.

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Alternate solution

The graph suggests an alternate solution. Since the T-shirt is caught at a height of 43 ft, this could be the maximum height the T-shirt reaches. That will occur at about t=0.564 seconds. Then the range of the function is 3–43 feet.

3x^3-x^2+6x-9

step-by-step explanation:

it's 15