A block of mass m is initially moving to the right on a horizontal frictionless surface at a speed v. It then compresses a spring of spring constant k. At the instant when the kinetic energy
of the block is equal to the potential energy of the spring, the spring is compressed a distance of:

The distance that the spring compresses is:

Explanation:

Kinetic and Elastic Potential Energy

The kinetic energy of an object of mass m traveling at a speed v is:

The elastic potential energy of a spring of constant k that compresses a distance x is:

The block of mass m is moving at a speed v when compresses a spring of constant k. The kinetic energy will eventually transform into elastic energy, but before that, both energies will be equal. It happens when:

Simplifying:

Dividing by k:

Taking square root:

The distance that the spring compresses is

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