A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground,away from the wall at the rate of 10 cm/s. How fast is the angle between the ladder and the ground decreasing when the foot of the ladder is 2 m away from the wall?

Options

A. `1/10` rad/s
B. `1/20` rad/s
C. 20 rad/s
D. 10 rad/s

Explanation

To solve the problem step by step, we will use the concepts of trigonometry and derivatives.

### Step 1: Understand the Problem
We have a ladder of length 5 m leaning against a wall. The bottom of the ladder is pulled away from the wall at a rate of 10 cm/s (which we will convert to meters). We need to find out how fast the angle between the ladder and the ground is decreasing when the foot of the ladder is 2 m away from the wall.

### Step 2: Draw a Diagram
Draw a right triangle where:
- The vertical side represents the height of the ladder against the wall (let's denote it as \( y \)).
- The horizontal side represents the distance from the wall to the foot of the ladder (denote it as \( x \)).
- The hypotenuse is the ladder itself, which is 5 m long.

### Step 3: Establish Relationships
From the right triangle, we can use the Pythagorean theorem:
\[ x^2 + y^2 = 5^2 \]
This simplifies to:
\[ x^2 + y^2 = 25 \]

### Step 4: Differentiate with Respect to Time
We need to differentiate the equation with respect to time \( t \):
\[ \frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(25) \]
Using the chain rule, we get:
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]
Dividing through by 2 gives:
\[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]

### Step 5: Substitute Known Values
We know:
- \( \frac{dx}{dt} = 10 \text{ cm/s} = 0.1 \text{ m/s} \)
- When \( x = 2 \text{ m} \), we can find \( y \) using the Pythagorean theorem:
\[ y^2 = 25 - x^2 = 25 - 2^2 = 25 - 4 = 21 \implies y = \sqrt{21} \]

Substituting \( x \), \( y \), and \( \frac{dx}{dt} \) into the differentiated equation:
\[ 2 \cdot 2 \cdot 0.1 + 2 \cdot \sqrt{21} \cdot \frac{dy}{dt} = 0 \]
This simplifies to:
\[ 0.4 + 2\sqrt{21} \frac{dy}{dt} = 0 \]
Solving for \( \frac{dy}{dt} \):
\[ 2\sqrt{21} \frac{dy}{dt} = -0.4 \implies \frac{dy}{dt} = -\frac{0.4}{2\sqrt{21}} = -\frac{0.2}{\sqrt{21}} \]

### Step 6: Relate Angle and Ladder Position
Using trigonometry, we know:
\[ \cos \theta = \frac{x}{5} \]
Differentiating both sides with respect to \( t \):
\[ -\sin \theta \frac{d\theta}{dt} = \frac{1}{5} \frac{dx}{dt} \]
Thus:
\[ \frac{d\theta}{dt} = -\frac{\frac{1}{5} \frac{dx}{dt}}{\sin \theta} \]

### Step 7: Find \( \sin \theta \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \cos \theta = \frac{x}{5} = \frac{2}{5} \implies \cos^2 \theta = \frac{4}{25} \]
Thus:
\[ \sin^2 \theta = 1 - \frac{4}{25} = \frac{21}{25} \implies \sin \theta = \frac{\sqrt{21}}{5} \]

### Step 8: Substitute Values
Substituting \( \frac{dx}{dt} = 0.1 \) and \( \sin \theta = \frac{\sqrt{21}}{5} \):
\[ \frac{d\theta}{dt} = -\frac{\frac{1}{5} \cdot 0.1}{\frac{\sqrt{21}}{5}} = -\frac{0.1}{\sqrt{21}} \]

### Conclusion
The angle between the ladder and the ground is decreasing at a rate of:
\[ \frac{d\theta}{dt} = -\frac{0.1}{\sqrt{21}} \text{ radians per second} \]