Consider the position function for free-falling objects.
s(t) = −16t2 + v0t + s0 

A ball is thrown straight down from the top of a 407-foot building with an initial velocity of −19 feet per second. Give the corresponding position function s(t).

How many seconds does it take the ball to fall 102 feet?

What is its velocity (in ft/s) after falling 102 feet?

Given the position function for free-falling objects:

\[ s(t) = -16t^2 + v_0t + s_0 \]

where:
- \( s(t) \) is the position at time \( t \)
- \( v_0 \) is the initial velocity
- \( s_0 \) is the initial position

In this scenario, a ball is thrown straight down from the top of a 407-foot building with an initial velocity of \(-19\) feet per second. Thus, \( v_0 = -19 \) and \( s_0 = 407 \).

The position function becomes:

\[ s(t) = -16t^2 - 19t + 407 \]

### Finding the Time to Fall 102 Feet

We need to determine the time it takes for the ball to fall 102 feet. The position \( s(t) \) when the ball has fallen 102 feet from the top of the building is:

\[ 407 - 102 = 305 \]

So, we set \( s(t) = 305 \) and solve for \( t \):

\[ 305 = -16t^2 - 19t + 407 \]

Rearranging the equation, we get:

\[ -16t^2 - 19t + 102 = 0 \]

To solve this quadratic equation, we can use the quadratic formula:

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

where \( a = -16 \), \( b = -19 \), and \( c = 102 \). Plugging in these values:

\[ t = \frac{-(-19) \pm \sqrt{(-19)^2 - 4(-16)(102)}}{2(-16)} \]
\[ t = \frac{19 \pm \sqrt{361 + 6528}}{-32} \]
\[ t = \frac{19 \pm \sqrt{6889}}{-32} \]
\[ t = \frac{19 \pm 83}{-32} \]

This gives us two solutions:

\[ t = \frac{19 + 83}{-32} = \frac{102}{-32} = -3.1875 \]
\[ t = \frac{19 - 83}{-32} = \frac{-64}{-32} = 2 \]

Since time cannot be negative, the time it takes for the ball to fall 102 feet is \( t = 2 \) seconds.

### Finding the Velocity After Falling 102 Feet

The velocity of a free-falling object is given by the derivative of the position function:

\[ v(t) = \frac{d}{dt}[-16t^2 - 19t + 407] \]

\[ v(t) = -32t - 19 \]

After 2 seconds:

\[ v(2) = -32(2) - 19 = -64 - 19 = -83 \]

So, the velocity of the ball after falling 102 feet is \(-83\) feet per second.

### Summary

1. The position function is \( s(t) = -16t^2 - 19t + 407 \).
2. The time it takes for the ball to fall 102 feet is \( 2 \) seconds.
3. The velocity after falling 102 feet is \(-83\) feet per second.

Would you like more details or have any questions?

Here are some additional questions you might want to explore:
1. What is the velocity of the ball just before it hits the ground?
2. How long does it take for the ball to reach the ground?
3. What is the maximum height reached by the ball if it were thrown upwards instead?
4. How would air resistance affect the fall of the ball?
5. What is the position of the ball after 1 second?

**Tip:** Always double-check your initial conditions and units when solving physics problems to ensure accuracy.

Explore how to calculate the position, time, and velocity of a free-falling object thrown from a 407-foot building with an initial velocity of -19 feet per second. Includes step-by-step solutions for finding the time it takes to fall 102 feet and the velocity after falling 102 feet. Suitable for Grades 11-12.

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