To determine the equation that models the height of the baseball as a function of time, we can follow a standard approach for projectile motion under the influence of gravity.
1. General Form of the Equation:
The general equation for height [tex]\( h(t) \)[/tex] of an object in free fall under gravity is:
[tex]\[
h(t) = -\frac{1}{2}gt^2 + vt + h_0
\][/tex]
where:
- [tex]\( g \)[/tex] is the acceleration due to gravity (in this case, 32 ft/s², because [tex]\( g = 2 \times 16 ft/s² \)[/tex])
- [tex]\( v \)[/tex] is the initial velocity (upwards)
- [tex]\( h_0 \)[/tex] is the initial height
2. Given Data:
- Initial height [tex]\( h_0 = 5 \)[/tex] feet
- Reaches maximum height (vertex) of 43.5 feet
- Total time in the air is 3.2 seconds
3. Determine the Initial Velocity:
Since the maximum height is reached and then the ball falls back, we can use the fact that maximum height occurs at the time when the velocity equals zero. The time to reach the maximum height is half of the total time, so it is [tex]\( \frac{3.2}{2} = 1.6 \)[/tex] seconds.
4. Using the Vertex Point:
The height at the vertex (maximum height) can be related to the initial velocity. The equation for height at time [tex]\( t \)[/tex] when [tex]\( t = 1.6 \)[/tex] seconds is:
[tex]\[
h(1.6) = -\frac{1}{2} \times 32 \times (1.6)^2 + v \times 1.6 + 5
\][/tex]
Simplifying [tex]\( \frac{1}{2} \times 32 = 16 \)[/tex], we get:
[tex]\[
h(1.6) = -16 \times (1.6)^2 + 1.6v + 5
\][/tex]
And we know [tex]\( h(1.6) = 43.5 \)[/tex]:
[tex]\[
43.5 = -16 \times 2.56 + 1.6v + 5
\][/tex]
Solving for [tex]\( v \)[/tex]:
[tex]\[
43.5 = -40.96 + 1.6v + 5
\][/tex]
[tex]\[
43.5 = -35.96 + 1.6v
\][/tex]
[tex]\[
43.5 + 35.96 = 1.6v
\][/tex]
[tex]\[
79.46 = 1.6v
\][/tex]
[tex]\[
v = \frac{79.46}{1.6} = 49.6625 \approx 49.64 \text{ ft/s}
\][/tex]
5. Final Equation:
Substituting back into the general form:
[tex]\[
h(t) = -16 t^2 + 49.64 t + 5
\][/tex]
So, the equation modeling the height of the baseball is:
[tex]\[
h(t) = -16t^2 + 49.64t + 5
\][/tex]
The correct choice from the given options is:
[tex]\[
h(t) = -16 t^2 + 49.64 t + 5
\][/tex]
