During halftime of a soccer ​game, a sling shot launches​ T-shirts at the crowd. A​ T-shirt is launched from a height of 6 feet with an initial upward velocity of 88 feet per second. The​ T-shirt is caught 36 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?Question content area bottomPart 1The​ T-shirt takesenter your response here ​second(s) to reach its maximum height.​(Type an integer or a​ decimal.)

1 answer

To determine how long it takes for the T-shirt to reach its maximum height and what that maximum height is, we can use the equations of motion for an object under the influence of gravity. The height \( h(t) \) of the T-shirt at time \( t \) seconds can be modeled by the following quadratic equation:

\[ h(t) = -16t^2 + v_0t + h_0 \]

where:

  • \( v_0 \) is the initial velocity (88 ft/s),
  • \( h_0 \) is the initial height (6 ft),
  • \( -16t^2 \) is the effect of gravity (measured in feet per second squared, using the conversion factor for the height equation).

Plugging in the values, we have:

\[ h(t) = -16t^2 + 88t + 6 \]

Part 1: Time to reach maximum height

The time to reach the maximum height for a quadratic function \( ax^2 + bx + c \) can be found using:

\[ t = -\frac{b}{2a} \]

In our equation:

  • \( a = -16 \)
  • \( b = 88 \)

So:

\[ t = -\frac{88}{2 \cdot -16} = \frac{88}{32} = 2.75 \text{ seconds} \]

Part 2: Maximum height

Now, plug \( t = 2.75 \) back into the height formula to find the maximum height:

\[ h(2.75) = -16(2.75)^2 + 88(2.75) + 6 \] \[ = -16(7.5625) + 242 + 6 \] \[ = -121 + 242 + 6 \] \[ = 127 \text{ feet} \]

Part 3: Range of the function

To determine the range of the function, we need to identify the minimum and maximum heights that the T-shirt achieves.

  • The maximum height is found at \( t = 2.75 \) seconds and is \( 127 \) feet.
  • The T-shirt is caught at a height of \( 36 \) feet.

Now we need to find when it reaches \( 36 \) feet. Set \( h(t) = 36 \):

\[ -16t^2 + 88t + 6 = 36 \]

Rearranging gives us:

\[ -16t^2 + 88t + 6 - 36 = 0 \implies -16t^2 + 88t - 30 = 0 \]

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

  • \( a = -16 \)
  • \( b = 88 \)
  • \( c = -30 \)

Calculating the discriminant:

\[ b^2 - 4ac = 88^2 - 4(-16)(-30) \] \[ = 7744 - 1920 = 5824 \]

Now, calculate the roots:

\[ t = \frac{-88 \pm \sqrt{5824}}{2 \cdot -16} \] Approximate the square root:

\(\sqrt{5824} \approx 76.3\)

Thus:

\[ t = \frac{-88 \pm 76.3}{-32} \]

Calculating the two possible times:

\[ t_1 = \frac{-88 + 76.3}{-32} \approx \frac{-11.7}{-32} \approx 0.3656 \text{ seconds} \]

\[ t_2 = \frac{-88 - 76.3}{-32} \approx \frac{-164.3}{-32} \approx 5.14375 \text{ seconds} \]

Range of the function

Since the T-shirt is launched and caught, the range of the function is between the height of 6 feet (the initial height) and 127 feet (the maximum height). However, since it is caught at 36 feet, the range can be stated as from 6 feet up to 127 feet, noting that it is caught at 36 feet.

Thus, the final answers are:

  • It takes 2.75 seconds to reach maximum height.
  • The maximum height is 127 feet.
  • The range of the function is [6, 127] (considering the initial height and the maximum height).