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

Explanation:

Step1: Determine the value of \(a\)

Given that the acceleration due to gravity is \(- 16\mathrm{ft/s}^2\). In the formula \(h(t)=at^{2}+vt + h_{0}\), the coefficient \(a=\frac{1}{2}\times\) acceleration. So \(a=\frac{-16}{2}=-8\).

Step2: Find \(h_{0}\)

When \(t = 0\), from the graph \(h(0)=5\). Substituting \(t = 0\) into \(h(t)=at^{2}+vt + h_{0}\), we get \(h(0)=a\times0^{2}+v\times0+h_{0}\). So \(h_{0}=5\).

Step3: Find \(v\)

We know the vertex of the parabola \(h(t)=at^{2}+vt + h_{0}\) occurs at \(t=-\frac{v}{2a}\). The vertex is at \(t = 1\) (from the point \((1,21)\) on the graph). Since \(a=-8\), then \(1=-\frac{v}{2\times(-8)}\).

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Step4: Write the equation

Substitute \(a=-8\), \(v = 16\), and \(h_{0}=5\) into \(h(t)=at^{2}+vt + h_{0}\). We get \(h(t)=-8t^{2}+16t + 5\)

Answer:

\(h(t)=-8t^{2}+16t + 5\)