To determine how high above the ground the baseball is 2 seconds after it is thrown, we can use the given quadratic equation for height, which is [tex]\(h(t) = at^2 + vt + h_0\)[/tex], where:
- [tex]\(h(t)\)[/tex] is the height of the ball above the ground at time [tex]\(t\)[/tex]
- [tex]\(a\)[/tex] is the acceleration due to gravity
- [tex]\(v\)[/tex] is the initial velocity of the ball
- [tex]\(h_0\)[/tex] is the initial height from which the ball is thrown
- [tex]\(t\)[/tex] is the time after the ball is thrown
We are given the following values:
- The initial height ([tex]\(h_0\)[/tex]) is 4 feet.
- The initial velocity ([tex]\(v\)[/tex]) is 36 feet per second.
- The acceleration due to gravity ([tex]\(a\)[/tex]) is [tex]\(-16\)[/tex] feet per second squared (negative because it acts downward).
- The time ([tex]\(t\)[/tex]) is 2 seconds.
Now, substitute these values into the equation:
[tex]\[ h(2) = -16 \cdot (2)^2 + 36 \cdot 2 + 4 \][/tex]
First, calculate [tex]\(2^2\)[/tex]:
[tex]\[ (2)^2 = 4 \][/tex]
Next, multiply it by [tex]\(-16\)[/tex]:
[tex]\[ -16 \cdot 4 = -64 \][/tex]
Then, multiply the initial velocity by time:
[tex]\[ 36 \cdot 2 = 72 \][/tex]
Add these values together with the initial height:
[tex]\[ h(2) = -64 + 72 + 4 \][/tex]
Combine the terms:
[tex]\[ h(2) = 8 + 4 \][/tex]
[tex]\[ h(2) = 12 \][/tex]
Thus, 2 seconds after the ball is thrown, it is 12 feet above the ground.
So, the correct answer is [tex]\(12 \, \text{ft}\)[/tex].
