Let's solve the problem step by step:
### Part A: Finding the Quadratic Regression Equation
To express the height [tex]\( h(t) \)[/tex] of the object as a function of the number of seconds since it was thrown, we perform a quadratic regression using the given data points. The general form of the quadratic equation is:
[tex]\[ h(t) = at^2 + bt + c \][/tex]
After performing the quadratic regression with the given data points:
[tex]\[ (t, h(t)) = \{(0.5, 78.125), (1, 90.8), (1.5, 101.025), (2, 108.8), (2.5, 114.125), (3, 117)\} \][/tex]
We find the coefficients:
[tex]\[ a = 63.0 \][/tex]
[tex]\[ b = 32.7 \][/tex]
[tex]\[ c = -4.9 \][/tex]
Thus, the height of the object as a function of time [tex]\( t \)[/tex] is:
[tex]\[ h(t) = 63.0t^2 + 32.7t - 4.9 \][/tex]
### Part B: Calculating Height at [tex]\( t = 2.8 \)[/tex] seconds
Using the regression equation [tex]\( h(t) \)[/tex], we can find the height of the object at [tex]\( t = 2.8 \)[/tex] seconds:
[tex]\[ h(2.8) = 63.0(2.8)^2 + 32.7(2.8) - 4.9 \][/tex]
[tex]\[ h(2.8) = 580.580 \][/tex]
Therefore, the height of the object 2.8 seconds after it is thrown is:
[tex]\[ 580.580 \, \text{meters} \][/tex]
### Part C: Calculating the Time to Reach a Height of 26 Meters
Using the regression equation [tex]\( h(t) \)[/tex], we want to find the time [tex]\( t \)[/tex] when the height [tex]\( h(t) \)[/tex] is 26 meters. We set up the equation:
[tex]\[ 63.0t^2 + 32.7t - 4.9 = 26 \][/tex]
This simplifies to:
[tex]\[ 63.0t^2 + 32.7t - 30.9 = 0 \][/tex]
Solving this quadratic equation for [tex]\( t \)[/tex], we find:
[tex]\[ t = 0.487 \][/tex]
Therefore, it will take approximately:
[tex]\[ 0.487 \, \text{seconds} \][/tex]
for the object to reach a height of 26 meters.
### Summary of Results:
- Part A: The height function is [tex]\( h(t) = 63.0t^2 + 32.7t - 4.9 \)[/tex].
- Part B: The height of the object at [tex]\( t = 2.8 \)[/tex] seconds is [tex]\( 580.580 \)[/tex] meters.
- Part C: The time for the object to reach a height of 26 meters is [tex]\( 0.487 \)[/tex] seconds.
