To solve for the values of [tex]\( t \)[/tex] when the ball's height [tex]\( h \)[/tex] is 16 feet, follow these steps:
1. Set the Height Equation Equal to 16:
The height [tex]\( h \)[/tex] as a function of time [tex]\( t \)[/tex] is given by:
[tex]\[
h = 4 + 29t - 16t^2
\][/tex]
Set [tex]\( h \)[/tex] to 16:
[tex]\[
16 = 4 + 29t - 16t^2
\][/tex]
2. Rearrange the Equation:
Move all terms to one side to set the equation to 0:
[tex]\[
0 = -16t^2 + 29t + 4 - 16
\][/tex]
Simplify the equation:
[tex]\[
0 = -16t^2 + 29t - 12
\][/tex]
3. Solve the Quadratic Equation:
This is a standard quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 29 \)[/tex], and [tex]\( c = -12 \)[/tex].
Use the quadratic formula:
[tex]\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Substituting in the coefficients:
[tex]\[
t = \frac{-29 \pm \sqrt{29^2 - 4(-16)(-12)}}{2(-16)}
\][/tex]
Simplifying inside the square root:
[tex]\[
t = \frac{-29 \pm \sqrt{841 - 768}}{-32}
\][/tex]
Further simplifying the square root:
[tex]\[
t = \frac{-29 \pm \sqrt{73}}{-32}
\][/tex]
4. Calculate the Two Possible Values of [tex]\( t \)[/tex]:
Solving for the two values of [tex]\( t \)[/tex]:
[tex]\[
t_1 = \frac{-29 + \sqrt{73}}{-32}
\][/tex]
and
[tex]\[
t_2 = \frac{-29 - \sqrt{73}}{-32}
\][/tex]
5. Determine the Numerical Values:
Calculate the values to the nearest hundredth:
[tex]\[
t_1 \approx 0.64
\][/tex]
[tex]\[
t_2 \approx 1.17
\][/tex]
So, the values of [tex]\( t \)[/tex] when the ball's height is 16 feet are:
[tex]\[
t = 0.64 \text{ seconds or } t = 1.17 \text{ seconds}
\][/tex]
