To determine which equation models the height of a cricket's jump, we can utilize the general form of the height function for vertical motion, which is given by:
[tex]\[ h(t) = -16t^2 + v_0 t + h_0 \][/tex]
Here, [tex]\(v_0\)[/tex] represents the initial vertical velocity, and [tex]\(h_0\)[/tex] represents the initial height from which the jump is made.
Given:
- The initial vertical velocity of the cricket, [tex]\(v_0\)[/tex], is [tex]\(14 \, \text{ft/s}\)[/tex].
- The initial height, [tex]\(h_0\)[/tex], is [tex]\(0 \)[/tex] feet (since we assume the jump starts from ground level).
We need to substitute these values into the general equation.
1. Substitute [tex]\(v_0 = 14\)[/tex] (initial vertical velocity) and [tex]\(h_0 = 0\)[/tex] (initial height) to get:
[tex]\[ h(t) = -16t^2 + 14t + 0 \][/tex]
Simplifying this, we get:
[tex]\[ h(t) = -16t^2 + 14t \][/tex]
2. Now, let's compare this resulting equation with the given options:
- [tex]\( h(t) = -16 t^2 + v_0 t + h_0 \quad \text{is written generically but we can simplify.} \)[/tex]
- [tex]\( h(t) = -16 t^2 + v_0 t + 14 \)[/tex]
- [tex]\( h(t) = -16 t^2 + 14 t \)[/tex]
- [tex]\( h(t) = -16 t^2 + 14 t + 14 \)[/tex]
The correct equation modeling the height of the cricket's jump after [tex]\( t \)[/tex] seconds, based on the initial velocity and initial height provided, is:
[tex]\[ h(t) = -16 t^2 + 14 t \][/tex]
Therefore, the equation that correctly models the height of the cricket's jump is:
[tex]\[ h(t) = -16 t^2 + 14 t \][/tex]
So, the correct answer is:
[tex]\[ h(t) = -16 t^2 + 14 t \][/tex]
