Answer:
The ball is six feet above the ground after two seconds.
Step-by-step explanation:
The height of a projectile as a function of time is modeled by the function:
[tex]\displaystyle h(t) = -2.5t^2 + 5t + 6[/tex]
And we want to determine after how many seconds is the ball six feet above the ground.
In other words, we can let h(t) = 6 and solve for t. This yields:
[tex]\displaystyle (6) = -2.5t^2 + 5t + 6[/tex]
Solve for t:
[tex]\displaystyle \begin{aligned}6 &= -2.5t^2 + 5t + 6 \\ 0 &= -2.5t^2 + 5t \\ 2.5t^2 - 5t &= 0 \\ 2.5t(t-2) &= 0 \end{aligned}[/tex]
By the Zero Product Property:
[tex]2.5t = 0\text{ or } t - 2 = 0[/tex]
Hence:
[tex]t = 0\text{ or } t = 2[/tex]
In conclusion, the ball is six feet above the ground after two seconds.
