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The problem describes a ball thrown vertically upward with an initial velocity of 32 feet per second, and its height \( s \) above the ground after \( t \) seconds is given by the equation:

\[
s = 32t - 16t^2
\]

We are asked to solve two parts:

**(a)** At what time will the ball strike the ground?

**(b)** For how long is the ball more than 12 feet above the ground?

Let's start by solving part (a).

### Part (a): Time when the ball strikes the ground
When the ball strikes the ground, \( s = 0 \). So, we set the equation equal to zero:

\[
0 = 32t - 16t^2
\]

Rearranging the equation:

\[
16t^2 = 32t
\]

Dividing both sides by 16:

\[
t^2 = 2t
\]

Now, factor the equation:

\[
t(t - 2) = 0
\]

Thus, \( t = 0 \) or \( t = 2 \).

Since \( t = 0 \) corresponds to the initial time (when the ball was thrown), the ball strikes the ground at \( t = 2 \) seconds.

### Part (b): Time when the ball is more than 12 feet above the ground
We now solve for when the ball is more than 12 feet above the ground, i.e., when \( s > 12 \). Using the height equation:

\[
32t - 16t^2 > 12
\]

Subtract 12 from both sides:

\[
32t - 16t^2 - 12 > 0
\]

Divide the entire inequality by 4 to simplify:

\[
8t - 4t^2 - 3 > 0
\]

We now solve the quadratic inequality \( -4t^2 + 8t - 3 > 0 \) by first solving the corresponding quadratic equation:

\[
-4t^2 + 8t - 3 = 0
\]

Using the quadratic formula:

\[
t = \frac{-8 \pm \sqrt{8^2 - 4(-4)(-3)}}{2(-4)}
\]
\[
t = \frac{-8 \pm \sqrt{64 - 48}}{-8}
\]
\[
t = \frac{-8 \pm \sqrt{16}}{-8}
\]
\[
t = \frac{-8 \pm 4}{-8}
\]

So, the two solutions are:

\[
t = \frac{-8 + 4}{-8} = \frac{-4}{-8} = \frac{1}{2}, \quad t = \frac{-8 - 4}{-8} = \frac{-12}{-8} = \frac{3}{2}
\]

Thus, the ball is more than 12 feet above the ground between \( t = \frac{1}{2} \) and \( t = \frac{3}{2} \) seconds.

### Summary:
- **(a)** The ball strikes the ground at \( t = 2 \) seconds.
- **(b)** The ball is more than 12 feet above the ground between \( t = \frac{1}{2} \) seconds and \( t = \frac{3}{2} \) seconds.

Would you like more details on any of these steps? Here are some follow-up questions to consider:

1. How do you derive the time of flight from a quadratic equation in general?
2. What is the significance of the initial velocity in this type of motion problem?
3. How can we calculate the maximum height the ball reaches?
4. Why do we ignore negative values for time in motion problems like this?
5. How can we adjust the equation if the ball were thrown from a height above the ground?

**Tip**: Always check the discriminant (the part under the square root) when solving quadratic equations. It tells you how many solutions to expect!

A ball is thrown vertically upward with an initial velocity of 32 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s = 32t - 16t^2. (a) At what time t will the ball strike the ground? (b) For what time t is the ball more than 12 feet above the ground?

Learn how to solve a quadratic motion problem where a ball is thrown vertically upward with an initial velocity of 32 feet per second. Find the time it hits the ground and when it is more than 12 feet above the ground. Includes detailed step-by-step solutions and key formulas for Grades 9-12.

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