Particle equation of motion with arbitrary parameterization vary the point particle action (5.15) to find a manifestly reparameterization invariant form of the free particle equation of motion. dxи dx" dr — прv dr dт сts s — —тс (5.15 т;

i don't know that i agree with that but what's the question ?

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

the answer is 100. i just did this lesson and got a 100% on it. : )

-fishtank