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can you solve 94, 95, & 96 and provide step by step to get the answers

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.



Can you solve 94, 95, & 96 and provide step by step to get the answers?

Solve step-by-step problems involving projectile motion. Learn how to calculate velocity and position for free-falling objects, including solving quadratic equations for time and analyzing the velocity function. Suitable for high school students in grades 11-12.

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