A bullet initially moving with a velocity 20 m `s^(-1)` strikes a target and comes to rest after penetrating a distance 10 cm in the target. Calculate the retardation caused by the target.

Explanation

To solve the problem of calculating the retardation caused by the target when a bullet strikes it, we can follow these steps:

### Step 1: Identify the known values
- Initial velocity of the bullet, \( u = 20 \, \text{m/s} \)
- Final velocity of the bullet, \( v = 0 \, \text{m/s} \) (since it comes to rest)
- Distance penetrated into the target, \( s = 10 \, \text{cm} = 0.1 \, \text{m} \) (convert centimeters to meters)

### Step 2: Use the kinematic equation
We can use the kinematic equation that relates initial velocity, final velocity, acceleration (or retardation), and distance:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) = final velocity
- \( u \) = initial velocity
- \( a \) = acceleration (which will be negative in this case since it's retardation)
- \( s \) = distance

### Step 3: Substitute the known values into the equation
Substituting the known values into the equation:
\[ 0^2 = (20)^2 + 2a(0.1) \]
This simplifies to:
\[ 0 = 400 + 0.2a \]

### Step 4: Solve for acceleration \( a \)
Rearranging the equation to solve for \( a \):
\[ 0.2a = -400 \]
\[ a = \frac{-400}{0.2} \]
\[ a = -2000 \, \text{m/s}^2 \]

### Step 5: Determine the retardation
Retardation is defined as the negative of acceleration:
\[ \text{Retardation} = -a = -(-2000) = 2000 \, \text{m/s}^2 \]

### Final Answer
The retardation caused by the target is \( 2000 \, \text{m/s}^2 \).

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