Question

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

Answer 1

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