Question

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

is modeled by the function h(t) = -2.5t2 + 5t +
6 where t is the number of seconds after the
projectile was launched and h is the height in feet.
After how many seconds is the ball 6 feet above
the ground? Round the nearest hundredth.

Answer 1

The ball is six feet above the ground after two seconds.

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.