A ball is thrown vertically upward from the top of a building 128 feet tall with an initial velocity of 112 feet per second.  The distance s​ (in feet) of the ball from the ground after t seconds is s equals 128 plus 112 t minus 16 t squared .
​(a) After how many seconds does the ball strike the​ ground?
​(b) After how many seconds will the ball pass the top of the building on its way​ down?
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Part 1
​(a) After how many seconds does the ball strike the​ ground?
t =

To solve the problem, we need to analyze the given equation for the position of the ball \( s(t) \) with respect to time \( t \):

\[
s(t) = 128 + 112t - 16t^2
\]

### Part (a): When does the ball strike the ground?
The ball strikes the ground when its distance from the ground \( s(t) \) is zero. So, we set the equation equal to zero and solve for \( t \):

\[
0 = 128 + 112t - 16t^2
\]

This is a quadratic equation in the standard form:

\[
-16t^2 + 112t + 128 = 0
\]

To solve for \( t \), we can use the quadratic formula:

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

Where \( a = -16 \), \( b = 112 \), and \( c = 128 \).

Let's calculate:

\[
t = \frac{-112 \pm \sqrt{112^2 - 4(-16)(128)}}{2(-16)}
\]

\[
t = \frac{-112 \pm \sqrt{12544 + 8192}}{-32}
\]

\[
t = \frac{-112 \pm \sqrt{20736}}{-32}
\]

\[
t = \frac{-112 \pm 144}{-32}
\]

This gives us two possible solutions:

1. \( t_1 = \frac{-112 + 144}{-32} = \frac{32}{-32} = -1 \) (not physically meaningful since time cannot be negative)
2. \( t_2 = \frac{-112 - 144}{-32} = \frac{-256}{-32} = 8 \)

Thus, the ball strikes the ground after \( t = 8 \) seconds.

### Part (b): When does the ball pass the top of the building on its way down?
The ball passes the top of the building when \( s(t) = 128 \) for the second time. Set the equation equal to 128 and solve for \( t \):

\[
128 = 128 + 112t - 16t^2
\]

Simplify the equation:

\[
0 = 112t - 16t^2
\]

Factor out \( t \):

\[
0 = t(112 - 16t)
\]

This gives us two solutions:

1. \( t = 0 \) (when the ball was initially at the top of the building)
2. \( 112 - 16t = 0 \), which simplifies to \( t = \frac{112}{16} = 7 \)

Thus, the ball passes the top of the building again after \( t = 7 \) seconds on its way down.

### Final Answers:
- **(a)** The ball strikes the ground after **8 seconds**.
- **(b)** The ball passes the top of the building on its way down after **7 seconds**.

Would you like to know more details or have any questions?

Here are 5 related questions:
1. What is the maximum height the ball reaches?
2. How long does it take for the ball to reach its maximum height?
3. What is the velocity of the ball when it strikes the ground?
4. How can we find the height of the ball at any given time?
5. How would the problem change if the initial velocity was different?

**Tip:** When solving quadratic equations in physics, always check the physical meaning of the solutions, as some roots may not be applicable.

Explore a projectile motion problem involving a ball thrown vertically upward from a building. Learn how to solve for the time it takes to strike the ground and pass the building's top, with detailed step-by-step solutions and explanations. Suitable for high school students studying quadratic equations and projectile motion in physics.

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