Question

a projectile is shot into the air. the altitude of the projectile after t seconds is modeled by the function graphed. time t is measured in feet.the projectile is $\boldsymbol{square}$ feet above the ground before launch.the projectile hits the ground after $\boldsymbol{square}$ seconds have passed.the projectile reaches a $\boldsymbol{square}$ height of $\boldsymbol{square}$ feet after being in the air for $\boldsymbol{square}$ seconds.the projectile is going up on from $\boldsymbol{square}$ and going down from $\boldsymbol{square}$

Explanation:

1. Before launch, time $t=0$. Use the point $(0, 5)$ from the graph.
2. The projectile hits the ground when height $h(t)=0$. Use the point $(25.93, 0)$ from the graph.
3. The maximum height is the vertex of the parabola, given by point $P(12, 19.4)$.
4. The projectile rises until it reaches the vertex, then falls afterward.

Answer:

The projectile is $\boldsymbol{5}$ feet above the ground before launch.
The projectile hits the ground after $\boldsymbol{25.93}$ seconds have passed.
The projectile reaches a maximum height of $\boldsymbol{19.4}$ feet after being in the air for $\boldsymbol{12}$ seconds.
The projectile is going up from $\boldsymbol{0}$ seconds and going down from $\boldsymbol{12}$ seconds to $\boldsymbol{25.93}$ seconds.