When you stand quietly, you pivot back and forth a very small amount about your ankles. It's easier to maintain stability if you tip slightly forward, so that your center of gravity is slightly in front of your ankles. The 71 kg man shown in (Figure 1) is 1.8 m tall; his center of gravity is 1.0 m above the ground and, during quiet standing, is 5.5 cm in front of his ankles. What is the magnitude of the gravitational torque about his ankles?

this isn't the answer but they're the graphs

change tension

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

amplitude

hope this