a mass of 2kg attached to a spiral spring to execute a SHM with amplitude of 0.12m its kinetic energy is 0.38J with its displacement is 0.07m calculate its frequency

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

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr