To determine which quadratic equation models the situation correctly, let's break down the information given and use the general form of a quadratic equation. The quadratic equation that models the height [tex]\( h(t) \)[/tex] of a projectile as a function of time [tex]\( t \)[/tex] is:
[tex]\[ h(t) = at^2 + vt + h_0 \][/tex]
where
- [tex]\( a \)[/tex] is the acceleration due to gravity (halved for the quadratic formula),
- [tex]\( v \)[/tex] is the initial velocity,
- [tex]\( h_0 \)[/tex] is the initial height.
### Given information:
1. Initial height [tex]\( h_0 = 3 \)[/tex] feet
2. Initial velocity [tex]\( v = 50 \)[/tex] feet per second
3. Acceleration due to gravity [tex]\( a = -16 \)[/tex] feet per second squared
### Step-by-Step Solution:
1. Identify the acceleration term [tex]\( a \)[/tex]:
[tex]\[
a = \frac{\text{Acceleration}}{2}
\][/tex]
Since the acceleration due to gravity is [tex]\(-16\)[/tex] feet per second squared:
[tex]\[
a = -16
\][/tex]
2. Substitute into the quadratic equation:
[tex]\[
h(t) = -16t^2 + 50t + 3
\][/tex]
Thus, the correct quadratic equation that models the height of the softball over time is:
[tex]\[ h(t) = -16t^2 + 50t + 3 \][/tex]
This corresponds to the second option provided:
[tex]\[ \boxed{h(t) = -16 t^2 + 50 t + 3} \][/tex]
