Question

22. The height of a projectile as a function of time

Answer 1

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:

$\displaystyle h(t) = -2.5t^2 + 5t + 6$

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:

$\displaystyle (6) = -2.5t^2 + 5t + 6$

Solve for t:

$\displaystyle \begin{aligned}6 &= -2.5t^2 + 5t + 6 \\ 0 &= -2.5t^2 + 5t \\ 2.5t^2 - 5t &= 0 \\ 2.5t(t-2) &= 0 \end{aligned}$

By the Zero Product Property:

$2.5t = 0\text{ or } t - 2 = 0$

Hence:

$t = 0\text{ or } t = 2$

In conclusion, the ball is six feet above the ground after two seconds.