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?
$h(t)=at^{2}+vt + h_{0}$

Explanation:

Step1: Determine the value of \(a\)

Given the acceleration due to gravity \(a=- 16\) (since \(h(t)=at^{2}+vt + h_{0}\) and acceleration is given as \(-16\ ft/s^{2}\))

Step2: Use the point \((0,5)\) to find \(h_{0}\)

Substitute \(t = 0\) and \(h(0)=5\) into \(h(t)=at^{2}+vt + h_{0}\).
We get \(h(0)=a\times0^{2}+v\times0+h_{0}\), so \(h_{0}=5\)

Step3: Use the point \((1,21)\) to find \(v\)

Substitute \(a=-16\), \(h_{0} = 5\), \(t = 1\) and \(h(1)=21\) into \(h(t)=at^{2}+vt + h_{0}\)

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Answer:

\(h(t)=-16t^{2}+42t + 5\)