Question

the graph below shows the height of a projectile t seconds after it is launched. if acceleration due to gravity is -16 ft/s², which equation models the height of the projectile correctly?
h(t) = at² + vt + h₀
o h(t) = -16t² + 42t + 5
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Explanation:

Step1: Substitute the points into the equation

Given \( h(t)=at^{2}+vt + h_{0}\).
When \(t = 0\), \(h(0)=5\), so \(h_{0}=5\).
When \(t = 1\), \(h(1)=a\times1^{2}+v\times1 + 5=21\), which simplifies to \(a + v=16\).
When \(t = 2.15\), \(h(2.15)=a\times(2.15)^{2}+v\times2.15 + 5 = 0\).
Since \(v = 16 - a\), substitute \(v\) into \(a\times(2.15)^{2}+v\times2.15 + 5 = 0\):

$$
LATEXBLOCK0
$$

If \(a=-16\), then \(v = 16-(-16)=32\).

Step2: Write the equation

Substitute \(a=-16\), \(v = 32\), and \(h_{0}=5\) into \(h(t)=at^{2}+vt + h_{0}\), we get \(h(t)=-16t^{2}+32t + 5\).

Answer:

\(h(t)=-16t^{2}+32t + 5\)