To solve the problem, we will use calculus, specifically differentiation, to find the velocity of the object and locate the time and height at its maximum position.

The height function is given as: $s(t) = -16t^2 + 256t + 192$

Step 1: Find the Object's Velocity at $t = 9$

The velocity of the object is the derivative of the position function $s(t)$ with respect to time $t$: $v(t) = s'(t) = \frac{d}{dt}(-16t^2 + 256t + 192)$ Differentiating, we get: $v(t) = -32t + 256$

Now, substitute $t = 9$: $v(9) = -32 \times 9 + 256 = -288 + 256 = -32 \, \text{feet per second}$

So, the velocity of the object at $t = 9$ seconds is $-32$ feet per second.

Step 2: Find the Maximum Height and When it Occurs

The maximum height occurs when the velocity is zero (the turning point in the function). Set $v(t) = 0$: $-32t + 256 = 0$ $t = \frac{256}{32} = 8 \, \text{seconds}$

Now, substitute $t = 8$ into the height function $s(t)$ to find the maximum height: $s(8) = -16(8)^2 + 256 \times 8 + 192$ $s(8) = -16 \times 64 + 2048 + 192$ $s(8) = -1024 + 2048 + 192 = 1216 \, \text{feet}$

Thus, the maximum height is $1216$ feet, and it occurs at $t = 8$ seconds.

Step 3: Find the Velocity When $s = 0$

To find the velocity when the object reaches the ground (when $s = 0$), set $s(t) = 0$: $-16t^2 + 256t + 192 = 0$ Divide the equation by $-16$: $t^2 - 16t - 12 = 0$ Solve this quadratic equation using the quadratic formula: $t = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1}$ $t = \frac{16 \pm \sqrt{256 + 48}}{2}$ $t = \frac{16 \pm \sqrt{304}}{2}$ $t = \frac{16 \pm 17.44}{2}$ This gives two solutions: $t \approx \frac{16 + 17.44}{2} \approx 16.72 \, \text{seconds} \quad \text{and} \quad t \approx \frac{16 - 17.44}{2} \approx -0.72 \, \text{(discard as time cannot be negative)}$

So, the object hits the ground at approximately $t = 16.72$ seconds.

Finally, substitute $t = 16.72$ into the velocity function: $v(16.72) = -32 \times 16.72 + 256$ $v(16.72) \approx -535.04 + 256 \approx -279.04 \, \text{feet per second}$

Summary of Results

1. Velocity at $t = 9$: $-32$ feet per second
2. Maximum Height: $1216$ feet at $t = 8$ seconds
3. Velocity when $s = 0$: Approximately $-279.04$ feet per second when $t \approx 16.72$ seconds

Would you like additional details on any of these steps or have further questions?

Related Questions

1. How do you interpret the negative velocity value?
2. What would the velocity be at the exact moment the object is launched?
3. How would air resistance affect the time and height calculations?
4. What other methods could confirm the maximum height reached by the object?
5. Could you find the average velocity of the object over a specific time interval?

Tip:

Remember, in vertical motion problems with constant acceleration, the derivative of position gives velocity, and the derivative of velocity gives acceleration, often simplifying problem-solving in such cases.