To model the situation of the softball being pitched, we need to form a quadratic equation that describes its height [tex]\( h(t) \)[/tex] at any time [tex]\( t \)[/tex]. The general form of the quadratic equation for the height in projectile motion is given by:
[tex]\[
h(t) = a t^2 + v t + h_0
\][/tex]
Here, [tex]\( a \)[/tex] represents the acceleration due to gravity, [tex]\( v \)[/tex] represents the initial velocity, and [tex]\( h_0 \)[/tex] represents the initial height.
Given the information:
- The initial velocity [tex]\( v \)[/tex] is 50 feet per second.
- The acceleration due to gravity [tex]\( a \)[/tex] is [tex]\( -16 \)[/tex] feet per second squared.
- The initial height [tex]\( h_0 \)[/tex] is 3 feet.
Let's translate these values into the quadratic equation format:
1. The acceleration [tex]\( a \)[/tex] is [tex]\( -16 \)[/tex], which will be multiplied by [tex]\( t^2 \)[/tex].
2. The initial velocity [tex]\( v \)[/tex] is [tex]\( 50 \)[/tex], which will be multiplied by [tex]\( t \)[/tex].
3. The initial height [tex]\( h_0 \)[/tex] is [tex]\( 3 \)[/tex], which will be added as a constant term.
So, plugging the given values into the equation [tex]\( h(t) = a t^2 + v t + h_0 \)[/tex], we get:
[tex]\[
h(t) = -16 t^2 + 50 t + 3
\][/tex]
Now let's examine the different options given in the problem:
1. [tex]\( h(t) = 50 t^2 - 16 t + 3 \)[/tex]
2. [tex]\( h(t) = -16 t^2 + 50 t + 3 \)[/tex]
3. [tex]\( 3 = -16 t^2 + 50 t + h_0 \)[/tex]
4. [tex]\( 3 = 50 t^2 - 16 t + h_0 \)[/tex]
Analyzing these:
- The first option is incorrect because it has the coefficients [tex]\( 50 \)[/tex] and [tex]\( -16 \)[/tex] in the wrong places.
- The third and fourth options are incorrect because they incorrectly place the initial height [tex]\( 3 \)[/tex] on the left side of the equation instead of incorporating it consistently on the right side as the constant term.
Therefore, the correct quadratic equation that models the situation is:
[tex]\[
h(t) = -16 t^2 + 50 t + 3
\][/tex]
This corresponds to option 2.
