A ball is dropped from a height of 5 m onto a sandy floor and penetrates the sand up to 10 cm before coming to rest. Find the retardation of the ball in and assuming it to be uniform.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the velocity of the ball just before it hits the sand.
The ball is dropped from a height of 5 meters. We can use the third equation of motion to find the final velocity (v) just before it hits the ground. The equation is:

\[ v^2 = u^2 + 2gh \]

Where:
- \( u = 0 \) (initial velocity, since the ball is dropped)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 5 \, \text{m} \) (height)

Substituting the values:

\[ v^2 = 0 + 2 \times 9.81 \times 5 \]
\[ v^2 = 98.1 \]
\[ v = \sqrt{98.1} \approx 9.9 \, \text{m/s} \]

### Step 2: Convert the penetration depth into meters.
The ball penetrates the sand up to 10 cm. We need to convert this into meters:

\[ D = 10 \, \text{cm} = 0.1 \, \text{m} \]

### Step 3: Use the third equation of motion to find the retardation.
When the ball penetrates the sand, it comes to rest. We can use the third equation of motion again:

\[ v'^2 = v^2 - 2aD \]

Where:
- \( v' = 0 \) (final velocity when the ball comes to rest)
- \( v = 9.9 \, \text{m/s} \) (velocity just before hitting the sand)
- \( D = 0.1 \, \text{m} \) (penetration depth)
- \( a \) is the retardation (which will be a positive value).

Substituting the values into the equation:

\[ 0 = (9.9)^2 - 2a(0.1) \]
\[ 0 = 98.01 - 0.2a \]

Rearranging the equation to solve for \( a \):

\[ 0.2a = 98.01 \]
\[ a = \frac{98.01}{0.2} \]
\[ a = 490.05 \, \text{m/s}^2 \]

### Step 4: Determine the retardation.
Since retardation is the negative acceleration, we can express it as:

\[ \text{Retardation} = -490.05 \, \text{m/s}^2 \]

### Final Answer:
The retardation of the ball while penetrating the sand is approximately \( 490.05 \, \text{m/s}^2 \) in the upward direction.

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