A golf ball is dropped from rest from a height of 8.30 m. It hits the pavement, then bounces back up, rising just 5.00 m before falling back down again. A boy then catches the ball when it is 1.10 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.

Now wait a minute, I already took this subject in 1954. I do not need practice. If you need help that is one thing but I see no sign of your attempts.

To find the total amount of time the ball is in the air, we need to consider the time it takes for the ball to fall from the initial height, the time it takes for the ball to reach its maximum height during the bounce, and the time it takes for the ball to fall from the maximum height to the height where the boy catches it.

Let's break down the problem step by step:

Step 1: Find the time it takes for the ball to fall from the initial height of 8.30 m to the pavement.

We can use the equation of motion:
h = ut + (1/2)gt^2

where h is the height, u is the initial velocity (0 in this case since the ball is dropped from rest), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

Rearranging the equation, we get:
t = sqrt(2h/g)

Substituting the values, we have:
t = sqrt(2 * 8.30 / 9.8) = 1.445 seconds (approximately)

So, the time for the ball to fall to the pavement is 1.445 seconds.

Step 2: Find the time it takes for the ball to reach its maximum height during the bounce.

Since the ball rises to a height of 5.00 m, we can use the same equation of motion. However, this time we consider the final height (0 m) as the reference point.

Using the equation:
h = ut + (1/2)gt^2

where h is the height, u is the initial velocity (0 when the ball reaches the maximum height), g is the acceleration due to gravity (-9.8 m/s^2 this time since the ball is moving upward), and t is the time.

Rearranging the equation, we get:
t = sqrt((-2h)/g)

Substituting the values for this case, we have:
t = sqrt((-2 * 5.00) / -9.8) = 1.020 seconds (approximately)

So, the time for the ball to reach its maximum height is 1.020 seconds.

Step 3: Find the time it takes for the ball to fall from the maximum height to the height where the boy catches it.

Similar to Step 1, we can use the equation of motion:
h = ut + (1/2)gt^2

This time, the initial height is 5.00 m, and we need to calculate the time it takes to fall to a height of 1.10 m.

Using the equation:
t = sqrt(2h/g)

Substituting the values, we have:
t = sqrt(2 * (5.00 - 1.10) / 9.8) = 0.853 seconds (approximately)

So, the time for the ball to fall from the maximum height to the height where the boy catches it is 0.853 seconds.

Step 4: Find the total time by adding up the times from each step.

Total time = time for falling to the pavement + time for reaching the maximum height + time for falling from the maximum height
= 1.445 + 1.020 + 0.853
= 3.318 seconds (approximately)

Hence, the total amount of time that the ball is in the air, from drop to catch, is approximately 3.318 seconds.