Pool players often pride themselves on their ability to impart a large speed to a pool ball. In the sport of billiards, event organizers often remove one of the rails on a pool table to allow players to measure the speed of their break shots (the opening shot of a game in which the player strikes a ball with his pool cue). With the rail removed, a ball can fly off the table. As the only participant with a physics background, they have placed you in charge of determining the speed of the players\' break shots.

The top of the pool table is 0.850 m from the floor. The placement of the tape is such that 0 m is aligned with the edge of the table. The winner of the competition wants to know if he has broken the world record for the break shot of 32 mph (about 14.3 m/s). If the winner\'s ball landed a distance 4.45 m from the table edge,

calculate his break shot speed. _____
At what speed did his pool ball hit the ground? ______

see your previous post, in the related questions below. Where did you get stuck?

To calculate the break shot speed and the speed at which the pool ball hit the ground, we can use the principles of projectile motion and the equations of kinematics.

1. Break Shot Speed:
The break shot is a projectile motion, where the ball is launched at an angle and follows a parabolic trajectory. In this case, we need to determine the initial velocity of the ball.

Given:
- Distance from the table edge (horizontal distance): x = 4.45 m
- Height of the table: h = 0.850 m

Let's assume the ball was launched at an angle of 45 degrees from the horizontal. We can use the following kinematic equation to calculate the initial velocity (v0) of the ball:
x = v0 * cos(theta) * t

Since the ball is launched from the ground:
h = v0 * sin(theta) * t - (1/2) * g * t^2

Where:
- v0 is the initial velocity magnitude
- theta is the launch angle
- t is the time of flight
- g is the acceleration due to gravity (approximately equal to 9.8 m/s^2)

For projectile motion, the time of flight (t) can be calculated using the equation:
t = (2 * v0 * sin(theta)) / g

Now, we can substitute this value of t into the equation for horizontal distance to solve for v0:
x = v0 * cos(theta) * [(2 * v0 * sin(theta)) / g]

Rearranging the equation, we get:
v0 = (x * g) / (2 * cos(theta) * sin(theta))

Plugging in the given values, we have:
v0 = (4.45 * 9.8) / (2 * cos(45°) * sin(45°))
= 6.967 m/s

Therefore, the break shot speed of the winner's pool ball is approximately 6.967 m/s.

2. Speed at Which the Pool Ball Hits the Ground:
To calculate the speed at which the pool ball hits the ground, we need to find the vertical component of its velocity at the moment it reaches the ground.

Using the equation we derived earlier for horizontal distance:
x = v0 * cos(theta) * t

Since the horizontal distance is given as x = 4.45 m, we can solve for the time of flight (t):
4.45 = (6.967 * cos(45°) * t)

Rearranging the equation, we get:
t = 4.45 / (6.967 * cos(45°))

Substituting the values, we have:
t = 0.892 s

Now, we can find the vertical component of the velocity (vy) at the moment the ball hits the ground using the equation:
vy = v0 * sin(theta) - g * t

Plugging in the values:
vy = 6.967 * sin(45°) - (9.8 * 0.892)
= 2.169 m/s

Therefore, the speed at which the pool ball hits the ground is approximately 2.169 m/s.