We are given the height function $s(t)$ of a stone thrown downward from a height of 1,200 feet:

$s(t) = 1200 - 104t - 16t^2$

(a) Find $s'(t)$, the velocity of the stone.

The derivative of $s(t)$, denoted as $s'(t)$, represents the velocity of the stone at any time $t$.

We differentiate $s(t)$ with respect to $t$:

$s'(t) = \frac{d}{dt}\left(1200 - 104t - 16t^2\right)$

• The derivative of the constant 1200 is 0.
• The derivative of $-104t$ is $-104$.
• The derivative of $-16t^2$ is $-32t$.

Thus, the velocity function is:

$s'(t) = -104 - 32t$

Now, we calculate the velocity at different times:

• $s'(0) = -104 - 32(0) = -104 \, \text{ft/s}$
• $s'(1) = -104 - 32(1) = -136 \, \text{ft/s}$
• $s'(2) = -104 - 32(2) = -168 \, \text{ft/s}$
• $s'(3) = -104 - 32(3) = -200 \, \text{ft/s}$
• $s'(4) = -104 - 32(4) = -232 \, \text{ft/s}$

So the velocities are:

• $s'(0) = -104 \, \text{ft/s}$
• $s'(1) = -136 \, \text{ft/s}$
• $s'(2) = -168 \, \text{ft/s}$
• $s'(3) = -200 \, \text{ft/s}$
• $s'(4) = -232 \, \text{ft/s}$

(b) When does the stone reach the ground, and how fast is it traveling at that point?

The stone reaches the ground when its height is 0, i.e., when $s(t) = 0$. We need to solve:

$1200 - 104t - 16t^2 = 0$

Rearrange the equation:

$16t^2 + 104t - 1200 = 0$

This is a quadratic equation. We can solve it using the quadratic formula:

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

For the equation $16t^2 + 104t - 1200 = 0$, the coefficients are:

• $a = 16$
• $b = 104$
• $c = -1200$

Now we calculate the time $t$.The two solutions for the time are $t = 6.0$ seconds and $t = -12.5$ seconds. Since negative time doesn't make sense in this context, the stone reaches the ground after 6 seconds.

Next, we find how fast the stone is traveling when it hits the ground. We substitute $t = 6$ into the velocity function $s'(t) = -104 - 32t$:

$s'(6) = -104 - 32(6) = -104 - 192 = -296 \, \text{ft/s}$

So, the stone is traveling at a velocity of $-296 \, \text{ft/s}$ when it hits the ground.

Summary:

• The stone reaches the ground at $t = 6$ seconds.
• The velocity when the stone hits the ground is $-296 \, \text{ft/s}$.

Would you like further details or have any questions? Here are some related questions for practice:

1. What is the stone’s velocity at $t = 5$ seconds?
2. What would be the initial height if the stone were thrown at $50 \, \text{ft/s}$ instead of $104 \, \text{ft/s}$?
3. How would the time to hit the ground change if the height were 1500 feet?
4. How can the equation be adjusted if the stone was thrown upward instead of downward?
5. What would the impact velocity be if the acceleration due to gravity was different?

Tip: When solving quadratic equations, always check the discriminant to see if real solutions exist.