To find all values of $t$ for which the height $h$ is 97 feet, we'll substitute $h = 97$ into the equation and solve for $t$:

$97 = 215t - 16t^2$

Rearrange the equation into standard form:

$-16t^2 + 215t - 97 = 0$

Now, we have a quadratic equation in the form $at^2 + bt + c = 0$, where:

• $a = -16$
• $b = 215$
• $c = -97$

Step 1: Solve Using the Quadratic Formula

The quadratic formula is: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute the values for $a$, $b$, and $c$:

$t = \frac{-215 \pm \sqrt{(215)^2 - 4(-16)(-97)}}{2(-16)}$

Step 2: Calculate the Discriminant

Calculate $b^2 - 4ac$: $(215)^2 - 4(-16)(-97) = 46225 - 6208 = 40017$

Step 3: Solve for $t$

Substitute the values back into the quadratic formula:

$t = \frac{-215 \pm \sqrt{40017}}{-32}$

Now we can evaluate this further:

$t = \frac{-215 \pm 200.04}{-32}$

This gives two solutions for $t$:

1. $t = \frac{-215 + 200.04}{-32} = \frac{-14.96}{-32} \approx 0.47$

2. $t = \frac{-215 - 200.04}{-32} = \frac{-415.04}{-32} \approx 12.97$

So, the values of $t$ for which the height is 97 feet are approximately: $t \approx 0.47 \text{ seconds} \quad \text{and} \quad t \approx 12.97 \text{ seconds}$

Would you like further details on any step, or do you have any questions?

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Here are some related questions that build on this problem:

1. What is the maximum height reached by the object?
2. For what values of $t$ will the object be at half the maximum height?
3. How long does it take for the object to return to a height of zero?
4. Can we determine the object's velocity at $t = 5$ seconds?
5. How does changing the initial velocity affect the height at a given time?

Tip: The discriminant $b^2 - 4ac$ tells you how many solutions a quadratic equation has: a positive value indicates two real solutions, zero means one real solution, and a negative value implies no real solutions.