The problem involves a ball thrown from an initial height of 3 feet with an initial upward velocity of 27 feet per second. The height $h$ of the ball after $t$ seconds is given by the equation:

$h = 3 + 27t - 16t^2$

We are tasked with finding the values of $t$ when the ball reaches a height of 13 feet. This involves solving the equation:

$13 = 3 + 27t - 16t^2$

Let's solve this quadratic equation step-by-step. First, we rearrange it as:

$0 = 3 + 27t - 16t^2 - 13$

Simplifying:

$0 = -16t^2 + 27t - 10$

Now, we will solve this quadratic equation using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = -16$, $b = 27$, and $c = -10$. Let's calculate the values of $t$.The two possible values of $t$ when the ball's height is 13 feet are approximately:

• $t_1 = 0.55$ seconds
• $t_2 = 1.14$ seconds

These are the times after the ball is thrown when it reaches a height of 13 feet.

Do you want more details on how these values were derived or have any questions?

Here are 5 related questions:

1. How is the discriminant used in solving quadratic equations?
2. What are the physical meanings of the two different times the ball reaches 13 feet?
3. How would the equation change if the initial velocity were different?
4. What is the significance of the negative coefficient in front of $t^2$?
5. How can we graph the height of the ball as a function of time?

Tip: The discriminant of a quadratic equation helps determine the number and nature of its solutions (real or complex).