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: Recall the acceleration - velocity - position relationship

The general equation for the height of a projectile is $h(t)=at^{2}+vt + h_{0}$, where $a$ is the acceleration, $v$ is the initial velocity, and $h_{0}$ is the initial height. Given that the acceleration due to gravity $a=- 16$ ft/s², the equation becomes $h(t)=-16t^{2}+vt + h_{0}$.

Step2: Use the initial - condition point $(0,5)$

When $t = 0$, $h(0)=5$. Substitute $t = 0$ into $h(t)=-16t^{2}+vt + h_{0}$:
$h(0)=-16(0)^{2}+v(0)+h_{0}=5$, so $h_{0}=5$.

Step3: Use the vertex point $(1,21)$

The vertex of a parabola $y = ax^{2}+bx + c$ has its $x$ - coordinate at $t=-\frac{b}{2a}$. For $h(t)=-16t^{2}+vt + 5$, $a=-16$ and $b = v$. Since the vertex is at $t = 1$, we have $1=-\frac{v}{2\times(-16)}$, which simplifies to $1=\frac{v}{32}$, so $v = 32$.

Step4: 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:

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