What is the frequency of a wave that passes a given point 57 times in 3.0 seconds?

The frequency of a wave is given by:

where

N is the number of complete cycles of the wave

t is the time taken

For the wave in this problem, we have 57 oscillations in 3.0 seconds, therefore N=57 and t=3.0 s. Substituting these numbers into the equation, we find

lol

explanation:

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