Let's break down the problems one by one and solve them step by step:

Problem 94:

A ball is thrown straight down from the top of a 220-foot building with an initial velocity of $-22 \text{ feet/second}$. We are tasked with:

1. Finding its velocity after 3 seconds.
2. Finding its velocity after falling 108 feet.

The position function is given as:

$s(t) = -16t^2 + v_0t + s_0$

Where:

• $v_0 = -22 \text{ ft/s}$ (initial velocity),
• $s_0 = 220 \text{ feet}$ (initial height),
• The acceleration due to gravity is approximated as $-16 \text{ ft/s}^2$.

(a) Velocity after 3 seconds:

The velocity function $v(t)$ is the derivative of the position function $s(t)$:

$v(t) = \frac{ds}{dt} = -32t + v_0$

Substitute $v_0 = -22$:

$v(t) = -32t - 22$

Now, find the velocity at $t = 3$:

$v(3) = -32(3) - 22 = -96 - 22 = -118 \text{ ft/s}$

So, after 3 seconds, the velocity of the ball is $-118 \text{ ft/s}$.

(b) Velocity after falling 108 feet:

To find the time when the ball has fallen 108 feet, we need to solve for $t$ when the ball’s height is $220 - 108 = 112$ feet:

$s(t) = -16t^2 - 22t + 220$

Set $s(t) = 112$:

$-16t^2 - 22t + 220 = 112$ $-16t^2 - 22t + 108 = 0$

Use the quadratic formula:

$t = \frac{-(-22) \pm \sqrt{(-22)^2 - 4(-16)(108)}}{2(-16)}$ $t = \frac{22 \pm \sqrt{484 + 6912}}{-32} = \frac{22 \pm \sqrt{7396}}{-32} = \frac{22 \pm 86}{-32}$

Taking the positive root (since time cannot be negative):

$t = \frac{22 + 86}{-32} = \frac{108}{-32} \approx -3.375 \text{ seconds}$

This value seems incorrect. We'll adjust this part after looking into the other problem step-by-step.