A bowling ball pendulum demonstration goes bad. The 8.0-lb bowling ball is raised to the height of the demonstrators nose, 160 cm above the ground, and releases the bowling ball. When the bowling ball is at the lowest point of the swing (36 inches above the ground), the knot slips and the ball flies forward through the air. How far forward will it fly from the release point before it strikes the ground?
Difference between 160 and 36 is 124, but I am not sure how to compute the rest of the problem
ah, 160 cm = 63 inches
63 - 36 = 27 inch fall
= 2.25 feet down fall
so how fast did it leave hand moving horizontally at t = 0?
m g h = (1/2) m u^2
u = sqrt (2 g h)
u = sqrt (2 * 32 * 2.25)
u = 12 feet/second horizontal speed
That will not change until we hit the pins if ever
now vertical fall from 3 feet up
3 = (1/2) g t^2 = 16 t^2
t = .433 seconds in the air
so
horizontal distance = u t
= 12 * .433 = 5.2 feet down the lane
To solve this problem, we can use the principle of conservation of energy. At the highest point of the pendulum swing, all of the potential energy is converted to kinetic energy. We can equate the potential energy at the release point with the kinetic energy at the lowest point to find the velocity of the bowling ball when it reaches the lowest point of the swing.
First, let's convert the given measurements into consistent units. The height of the ball at the release point is 160 cm. To convert it to meters, divide by 100. So, the height is 1.6 meters. The height at the lowest point is given as 36 inches, which is approximately 0.9144 meters.
Next, calculate the gravitational potential energy at the release point. The formula for potential energy is PE = m * g * h, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.
Potential energy at the release point (PE1) = 8.0 lb * (0.4536 kg/lb) * 9.8 m/s² * 1.6 m
Now that we have the potential energy at the release point, we can equate it to the kinetic energy at the lowest point. The formula for kinetic energy is KE = (1/2) * m * v², where m is the mass and v is the velocity.
Kinetic energy at the lowest point (KE2) = (1/2) * 8.0 lb * (0.4536 kg/lb) * v²
Since the mass of the bowling ball cancels out, we can equate PE1 to KE2 and solve for v.
8.0 lb * 0.4536 kg/lb * 9.8 m/s² * 1.6 m = (1/2) * 8.0 lb * 0.4536 kg/lb * v²
Simplifying the equation, we get:
9.8 m/s² * 1.6 m = (1/2) * v²
Solving for v, we find:
v = √(9.8 m/s² * 1.6 m * 2)
Now, you can calculate the velocity at the lowest point of the swing.
After the knot slips and the ball flies forward, it will continue to move horizontally with the same velocity. The horizontal distance traveled can be calculated using the equation d = v * t, where d is the distance, v is the velocity, and t is the time.
To find the time taken, we can use the second equation of motion: d = ut + (1/2) * a * t², where d is the distance (0.9144 m), u is the initial velocity (v), a is the acceleration (0 m/s² since there is no horizontal acceleration), and t is the time.
0.9144 m = v * t
Now, solve the equation for t in terms of v and substitute the value of v you calculated earlier.
Finally, substitute the values of v and t into the equation d = v * t to find the horizontal distance traveled.
So, by following these steps, you can find the distance forward the ball will fly from the release point before striking the ground.