Explanation:

To determine the speed of the ball at the instant it touches the cake surface and its retardation inside the cake, we can analyze the motion of the ball during its fall.

Given:

Mass of the ball = m

Initial height above the cake surface = H

Thickness of the cake = h

When the ball falls freely, it accelerates due to gravity (g) and reaches a certain velocity before it touches the cake surface. At this point, the ball has converted its potential energy into kinetic energy.

Using the principle of conservation of energy, we can equate the initial potential energy (mgh) to the final kinetic energy (1/2 mv^2), where v is the speed of the ball when it touches the cake surface.

mgh = 1/2 mv^2

We can cancel the mass (m) from both sides of the equation:

gh = 1/2 v^2

Simplifying, we get:

v = √(2gh)

This is the speed of the ball at the instant it touches the cake surface.

Once the ball penetrates the cake, it experiences a retarding force due to the resistance offered by the cake material. This force opposes the motion of the ball and leads to a negative acceleration or retardation.

The retardation inside the cake can be calculated using Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration:

F = ma

In this case, the force (F) is the retarding force, the mass (m) is the mass of the ball, and the acceleration (a) is the retardation inside the cake.

The force of retardation can be related to the speed (v) of the ball using the equation:

F = -bv

Where b is a constant that depends on the properties of the cake material.

Substituting these values into Newton's second law, we have:

-bv = ma

Since acceleration (a) is the derivative of velocity (v) with respect to time, we can rewrite the equation as:

-bv = m(dv/dt)

Rearranging the equation, we get:

dv/v = (-b/m) dt

Integrating both sides, we have:

∫ dv/v = ∫ (-b/m) dt

ln(v) = (-b/m) t + C

Where C is the constant of integration.

Exponentiating both sides, we get:

v = e^((-b/m) t + C)

Simplifying, we have:

v = Ae^(-bt/m)

Where A is the constant obtained by exponentiating C.

This equation describes the velocity of the ball inside the cake as a function of time. The negative sign indicates that the velocity decreases with time, representing the retardation.

Therefore, the speed of the ball at the instant it touches the cake surface is v = √(2gh), and the retardation inside the cake is given by v = Ae^(-bt/m).