Pool players often pride themselves on their ability to impart a large speed on 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 0.770 m from the floor. The placement of the tape is such that 0 m aligns to the edge of the table (as shown). 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). The ball landed a distance 5.15 m from the table edge, calculate his break shot speed.
At what speed did his pool ball hit the ground?
Well, well, well, it seems we have a record-breaking pool player here! Let's dive into the physics of this epic break shot, shall we?
To determine the speed at which the pool ball hit the ground, we can use some basic kinematic equations. The key is to find the time it took for the ball to travel 5.15 m horizontally, from the table to the landing spot.
First, we need to find the time it took for the ball to reach the ground vertically. We can use the equation:
h = (1/2)gt^2
Where:
h is the vertical distance from the table top to the ground (0.770 m).
g is the acceleration due to gravity (approximately 9.8 m/s^2).
t is the time we're looking for.
Rearranging the equation, we get:
t = sqrt(2h/g)
t = sqrt(2 * 0.770 m / 9.8 m/s^2)
After crunching the numbers, we find that t ≈ 0.156 seconds.
Now that we know how long it took for the ball to reach the ground vertically, we can calculate its vertical velocity using the equation:
v = gt
v = 9.8 m/s^2 * 0.156 s
v ≈ 1.529 m/s
So, the speed at which the pool ball hit the ground is approximately 1.529 m/s.
But you might be thinking, Clown Bot, how does this help us determine the break shot speed? Well, my friend, we just need to use a little trigonometry. Since the ball landed 5.15 m horizontally from the table edge, we can determine the time it took for the ball to travel that distance using:
v = d/t
Where:
v is the horizontal velocity (our target variable).
d is the horizontal distance traveled (5.15 m).
t is the time.
Rearranging the equation, we get:
t = d/v
Plugging in the values, we get:
t = 5.15 m / v
Now, we know that the total time of flight is the sum of the vertical time and the horizontal time:
Total time = t + t = 2t
So, we have:
Total time = 2t ≈ 2 * 0.156 seconds
Total time ≈ 0.312 seconds
Finally, we can find the break shot speed using:
v = d / t
Plugging in the values, we have:
v = 5.15 m / 0.312 s
After doing the math, we get:
v ≈ 16.506 m/s
So, it seems our record-breaking pool player's break shot speed was approximately 16.506 m/s. While it falls short of the world record, it's still quite an impressive feat!
Remember, folks, when it comes to physics, it's all about finding the right angles and shooting for the stars... or in this case, the pool table! Keep up the great work, pool player extraordinaire!
To calculate the speed at which the pool ball hit the ground, we can start by using the horizontal distance travelled by the ball (5.15 m) and the height of the table (0.770 m) to find the time it took for the ball to reach the ground using the equations of motion.
The horizontal distance traveled by the ball (x) is given as 5.15 m.
The height of the table (h) is given as 0.770 m.
We can use the equation for horizontal distance, x = v0*t, where v0 is the initial horizontal velocity of the ball and t is the time it takes to reach the ground.
Let's assume the ball was launched horizontally, therefore the initial vertical velocity, Vy0, is 0 m/s. The only force acting on the ball in the vertical direction is gravity.
Using the equation for vertical displacement, h = Vy0*t + (1/2)*g*t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2), we can solve for t.
Rearranging the equation, we have:
0.770 m = (1/2)*9.8 m/s^2 * t^2
Simplifying, we get:
t^2 = (2*0.770 m) / 9.8 m/s^2
t^2 = 0.15714 s^2
t ≈ 0.3966 s
Now, we can use the equation x = v0*t to calculate the initial horizontal velocity, v0.
Plugging in the values we know:
5.15 m = v0 * 0.3966 s
Solving for v0, we have:
v0 = 5.15 m / 0.3966 s
v0 ≈ 12.98 m/s
Therefore, the speed at which the ball hit the ground was approximately 12.98 m/s.
To calculate the speed at which the pool ball hit the ground, you can use the principles of projectile motion and the equation for vertical displacement.
First, let's define the variables:
- The vertical displacement, h, is the height of the table from the floor, given as 0.770 m.
- The horizontal displacement, x, is the distance the ball traveled before hitting the ground, given as 5.15 m.
- The acceleration due to gravity, g, is approximately 9.8 m/s^2.
Now, we can use the equation for vertical displacement during projectile motion:
h = (1/2) * g * t^2
However, to solve for the time of flight, t, we need to find the time it takes for the ball to hit the ground after being struck by the cue.
Using the horizontal displacement and the initial horizontal velocity, we can find the time of flight, t:
x = v_x * t
v_x is the initial horizontal velocity, which is the speed at which the ball was hit. This is what we are trying to find.
Now, let's solve for t:
t = x / v_x
Substituting the values of x and v_x:
t = 5.15 m / v_x
Now, substitute the expression for time, t, into the equation for vertical displacement and solve for v_x:
h = (1/2) * g * (x / v_x)^2
Rearranging the equation, we get:
v_x^2 = g * x / (2h)
Substituting the known values:
v_x^2 = 9.8 m/s^2 * 5.15 m / (2 * 0.770 m)
v_x^2 = 33.47 m^2/s^2
To find the speed at which the pool ball hit the ground, we take the square root of v_x^2:
v_x = sqrt(33.47 m^2/s^2)
v_x ≈ 5.789 m/s
Therefore, the speed at which the pool ball hit the ground is approximately 5.789 m/s.
a. h = 0.5g*t^2 = 0.77
4.9t^2 = 0.77
t^2 = 0.1571
t = 0.396 s. = Fall time.
Dx = Xo * t = 5.15 m.
Xo * 0.396 = 5.15
Xo = 13 m/s = Break speed.
b. V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*0.77 = 15.09
V = 3.88 m/s