You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.20 m. How much additional time will pass before the ball passes the tree branch on the way back down?

To solve this problem, we can break it into two steps:

1. **Find the time it takes for the ball to reach the height of 7.20 m on the way up.**
2. **Determine the additional time it takes for the ball to return to that height (on the way down) after reaching the maximum height.**

### Step 1: Find the time taken to reach the tree branch (7.20 m) on the way up.

We can use the following kinematic equation:

\[
h = v_0 t - \frac{1}{2} g t^2
\]

where:
- \( h = 7.20 \, \text{m} \) (height of the tree branch),
- \( v_0 = 15.0 \, \text{m/s} \) (initial velocity),
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( t \) is the time.

Rearranging the equation:

\[
0 = -\frac{1}{2} g t^2 + v_0 t - h
\]

Substituting the known values:

\[
0 = -\frac{1}{2}(9.8)t^2 + 15.0 t - 7.20
\]

\[
0 = -4.9 t^2 + 15.0 t - 7.20
\]

This is a quadratic equation. Solving for \( t \), we can use the quadratic formula:

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

where:
- \( a = -4.9 \),
- \( b = 15.0 \),
- \( c = -7.20 \).

### Step 2: Solve for \( t \).

Let me compute the solution for the quadratic equation.The two possible solutions for time \( t \) are approximately:

- \( t_1 = 0.60 \, \text{seconds} \) (on the way up),
- \( t_2 = 2.47 \, \text{seconds} \) (on the way down).

### Step 3: Determine the additional time before the ball passes the tree branch on the way down.

The additional time after the ball reaches its maximum height is the difference between these two times:

\[
\Delta t = t_2 - t_1 = 2.47 \, \text{seconds} - 0.60 \, \text{seconds} = 1.87 \, \text{seconds}
\]

Thus, the ball will take an additional **1.87 seconds** to pass the tree branch on the way back down.

Would you like more details or further clarification? 

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Here are 5 related questions to think about:

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 total time the ball spends in the air before returning to the ground?
4. How would the result change if the initial velocity were different?
5. What is the velocity of the ball when it passes the tree branch on the way down?

**Tip:** Quadratic equations often have two solutions—it's important to understand their physical meanings in motion problems!

Learn how to calculate the additional time for a ball, thrown with an initial velocity of 15.0 m/s, to pass a height of 7.20 m on its way back down. This physics problem involves kinematic equations and quadratic solutions, ideal for high school students in Grades 9-12.

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