Let's solve the problem step-by-step.
We are given:
- The initial height [tex]\( h_0 = 2 \)[/tex] feet.
- The initial velocity [tex]\( V_0 = 130 \)[/tex] feet per second.
The general equation for the projectile's height [tex]\( h(t) \)[/tex] as a function of time [tex]\( t \)[/tex] is given by:
[tex]\[ h(t) = -16t^2 + V_0 t + h_0 \][/tex]
We need to substitute the given values of [tex]\( V_0 \)[/tex] and [tex]\( h_0 \)[/tex] into this equation.
Step 1: Substitute [tex]\( V_0 = 130 \)[/tex] into the equation.
[tex]\[ h(t) = -16t^2 + 130t + h_0 \][/tex]
Step 2: Substitute [tex]\( h_0 = 2 \)[/tex] into the equation.
[tex]\[ h(t) = -16t^2 + 130t + 2 \][/tex]
So the equation that models the ball's height as a function of time is:
[tex]\[ h(t) = -16t^2 + 130t + 2 \][/tex]
Now, let's match this with the given options:
A. [tex]\( h(t) = -16t^2 + 2t + 130 \)[/tex]
- This equation incorrectly places the initial height and the initial velocity coefficients in the wrong order.
B. [tex]\( h(t) = -16t^2 - 130t + 2 \)[/tex]
- This equation incorrectly has a negative initial velocity term [tex]\( -130t \)[/tex].
C. [tex]\( h(t) = -16t^2 + 130t + 2 \)[/tex]
- This equation is correct and matches our derived equation.
D. [tex]\( h(t) = -16t^2 - 2t + 130 \)[/tex]
- This equation incorrectly places the initial height and the initial velocity coefficients in the wrong order and signs.
Therefore, the correct equation is:
[tex]\[ c. h(t) = -16t^2 + 130t + 2 \][/tex]
