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: Substitute points into equation

We know \(h(t)=at^{2}+vt + h_{0}\), and the acceleration \(a=- 16\). Substitute the points \((0,5)\), \((1,21)\) and \((2.15,0)\) into the equation. When \(t = 0\), \(h(0)=h_{0}\), so \(h_{0}=5\) since \(h(0)=5\).
When \(t = 1\), \(h(1)=-16\times1^{2}+v\times1 + 5\). Since \(h(1) = 21\), we have \(-16 + v+5=21\), which simplifies to \(v-11 = 21\), and then \(v=32\).

Step2: Write the final 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:

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