To solve this problem, we can break it into three phases:

1. Fall from 9.60 m to the pavement.
2. Bounce up to a height of 5.70 m and fall back down to the pavement.
3. Rise from pavement to 1.30 m and then be caught by the boy.

We'll use the following kinematic equation to find the time for each phase:

$h = \frac{1}{2} g t^2$

where:

• $h$ is the height (distance fallen or risen),
• $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity,
• $t$ is the time.

Step 1: Fall from 9.60 m

Using $h = \frac{1}{2} g t_1^2$, where $h = 9.60 \, \text{m}$,

$9.60 = \frac{1}{2} \times 9.8 \times t_1^2$

Solving for $t_1$:

$t_1 = \sqrt{\frac{2 \times 9.60}{9.8}}$

Step 2: Bounce to 5.70 m and fall back down

Since the motion is symmetric, the time to rise and fall will be the same. First, we calculate the time to rise to 5.70 m:

$5.70 = \frac{1}{2} \times 9.8 \times t_2^2$

The total time for the rise and fall is $2t_2$.

Step 3: Rise to 1.30 m after bounce

For the rise to 1.30 m:

$1.30 = \frac{1}{2} \times 9.8 \times t_3^2$

Now, let’s solve for $t_1$, $t_2$, and $t_3$.The total time that the golf ball is in the air, from the initial drop to when it is caught by the boy, is approximately $4.07 \, \text{seconds}$.

Do you need any further details or clarification?

Here are some related questions you might find interesting:

1. How would the total time change if air resistance was considered?
2. How does the maximum height after bouncing affect the time in the air?
3. What would happen if the ball didn't bounce as high after hitting the ground?
4. Can we determine the velocity of the ball at the moment the boy catches it?
5. How does the mass of the ball influence its motion if air resistance was not ignored?

Tip: The kinematic equations assume constant acceleration, making them ideal for problems involving gravity, as long as air resistance is negligible.