A 0.500-kg stone is moving in a vertical circular path attached to a string that is 75.0 cm long. The stone is moving around the path at a constant frequency of 1.50 rev/sec. At the moment the stone is overhead, the stone is released. The velocity of the stone when it leaves the circular path is

A. 5.55 m/s.
B. 7.07 m/s.
C. 7.75 m/s.
D. 8.35 m/s.
E. 9.00 m/s.

B. 7.07 m/s

Explanation:

The velocity of the stone when it leaves the circular path is its tangential velocity, , which is given by

where is the angular speed and is the radius of the circular path.

is given by

where is the frequency of revolution.

Thus

Using values from the question,

Note the conversion of 75 cm to 0.75 m

i think it's 35 meters

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft