How does the decrease in gravitational potential energy of a falling ball compare to its increase in kinetic energy? (Ignore air friction.)

for every decrease in potential energy, a kinetic energy proportional to 2gΔh is gained.

Explanation:

Let us consider a ball falling from its maximum height

For a body falling from its maximum height to a point p

change in height = Δh

The potential energy decrease is then proportional to

ΔPE = mgΔh

where

ΔPE is the decrease in kinetic energy

m is the mass of the ball

g is acceleration due to gravity

Δh is the change in height

For a body falling from its maximum height, the increase change in velocity

Δv = u + 2gΔh    (at maximum height u = 0)

where

u is the initial kinetic energy of the ball

Δv = 0 + 2gΔh

Δv = 2gΔh

The kinetic energy increases by

ΔKE = m(Δv)^2

but Δv = 2gΔh

therefore

ΔKE = m(2gΔh)^2 = 2m(gΔh)^2

comparing the increase in kinetic energy to the decrease in potential energy, we have

(2m(gΔh)^2)/(mgΔh) = 2gΔh

This means that for every decrease in potential energy, a kinetic energy proportional to 2gΔh is gained.

d is the correct answer

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