The problem in the image gives a quadratic equation for the height $h$ of an object thrown upward:

$h = -16t^2 + 75t + 16$

where:

• $h$ is the height after $t$ seconds,
• $t$ is the time in seconds.

1. When will the height be 22 feet?

To solve this, we need to set $h = 22$ and solve the quadratic equation:

$-16t^2 + 75t + 16 = 22$

Simplify the equation:

$-16t^2 + 75t + 16 - 22 = 0$

$-16t^2 + 75t - 6 = 0$

Now, use the quadratic formula:

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

where $a = -16$, $b = 75$, and $c = -6$.

$t = \frac{-75 \pm \sqrt{75^2 - 4(-16)(-6)}}{2(-16)}$

$t = \frac{-75 \pm \sqrt{5625 - 384}}{-32}$

$t = \frac{-75 \pm \sqrt{5241}}{-32}$

$t = \frac{-75 \pm 72.39}{-32}$

Now calculate both roots:

1. $t_1 = \frac{-75 + 72.39}{-32} = \frac{-2.61}{-32} = 0.0816$ seconds.
2. $t_2 = \frac{-75 - 72.39}{-32} = \frac{-147.39}{-32} = 4.61$ seconds.

Thus, the height will be 22 feet at approximately 0.08 seconds and 4.61 seconds.

2. When will the object reach the ground?

To find when the object reaches the ground, we set $h = 0$:

$-16t^2 + 75t + 16 = 0$

Use the quadratic formula again with $a = -16$, $b = 75$, and $c = 16$:

$t = \frac{-75 \pm \sqrt{75^2 - 4(-16)(16)}}{2(-16)}$

$t = \frac{-75 \pm \sqrt{5625 + 1024}}{-32}$

$t = \frac{-75 \pm \sqrt{6649}}{-32}$

$t = \frac{-75 \pm 81.53}{-32}$

Now calculate the roots:

1. $t_1 = \frac{-75 + 81.53}{-32} = \frac{6.53}{-32} = -0.20$ (discard this as time can't be negative),
2. $t_2 = \frac{-75 - 81.53}{-32} = \frac{-156.53}{-32} = 4.89$ seconds.

Thus, the object will reach the ground at approximately 4.89 seconds.

Would you like more details or have any questions?

Here are five follow-up questions you might consider:

1. How does the quadratic formula help solve projectile motion problems?
2. What would change if the initial speed or height of the object were different?
3. How can you interpret the two different times when the object reaches 22 feet?
4. What does the negative root for time physically represent in these problems?
5. How could you derive the maximum height of the object?

Tip: Always check for reasonable physical meanings of time values—negative time often has no real-world meaning in these contexts!