A man can throw a stone with initial speed of `10 m//s`. Find the maximum horizontal distance to which he can throw the stone in a room of height h for:
(i) `h = 2 m` and (ii) `h = 4 m`.

Answer

To solve the problem of finding the maximum horizontal distance a man can throw a stone in a room of height \( h \) with an initial speed of \( 10 \, \text{m/s} \), we will use the concepts of projectile motion.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We need to find the maximum horizontal distance (range) for two different heights: \( h = 2 \, \text{m} \) and \( h = 4 \, \text{m} \).

2. **Using the Maximum Height Formula:**
The maximum height \( h \) reached by a projectile is given by the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
where:
- \( u \) is the initial speed (10 m/s),
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)),
- \( \theta \) is the angle of projection.

3. **Finding \( \theta \) for \( h = 2 \, \text{m} \):**
Substitute \( h = 2 \, \text{m} \) into the formula:
\[ 2 = \frac{10^2 \sin^2 \theta}{2 \times 10} \]
Simplifying this gives:
\[ 2 = \frac{100 \sin^2 \theta}{20} \implies 2 = 5 \sin^2 \theta \implies \sin^2 \theta = \frac{2}{5} \]
Therefore:
\[ \sin \theta = \sqrt{\frac{2}{5}} \]

4. **Finding \( \cos \theta \):**
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{2}{5} = \frac{3}{5} \]
Thus:
\[ \cos \theta = \sqrt{\frac{3}{5}} \]

5. **Calculating the Range \( R \):**
The range \( R \) is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Using the double angle identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ R = \frac{10^2 \cdot 2 \cdot \sqrt{\frac{2}{5}} \cdot \sqrt{\frac{3}{5}}}{10} \]
Simplifying this:
\[ R = \frac{100 \cdot 2 \cdot \sqrt{\frac{6}{25}}}{10} = 20 \cdot \sqrt{\frac{6}{25}} = 4\sqrt{6} \, \text{m} \]

6. **Finding \( \theta \) for \( h = 4 \, \text{m} \):**
Substitute \( h = 4 \, \text{m} \) into the height formula:
\[ 4 = \frac{10^2 \sin^2 \theta}{2 \times 10} \]
Simplifying gives:
\[ 4 = 5 \sin^2 \theta \implies \sin^2 \theta = \frac{4}{5} \]
Therefore:
\[ \sin \theta = \sqrt{\frac{4}{5}} \]

7. **Finding \( \cos \theta \) for \( h = 4 \, \text{m} \):**
Using the identity:
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{4}{5} = \frac{1}{5} \]
Thus:
\[ \cos \theta = \sqrt{\frac{1}{5}} \]

8. **Calculating the Range \( R \) for \( h = 4 \, \text{m} \):**
Using the range formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Substituting:
\[ R = \frac{10^2 \cdot 2 \cdot \sqrt{\frac{4}{5}} \cdot \sqrt{\frac{1}{5}}}{10} \]
Simplifying gives:
\[ R = \frac{100 \cdot 2 \cdot \sqrt{\frac{4}{25}}}{10} = 20 \cdot \frac{2}{5} = 8 \, \text{m} \]

### Final Answers:
- For \( h = 2 \, \text{m} \), the maximum horizontal distance is \( 4\sqrt{6} \, \text{m} \).
- For \( h = 4 \, \text{m} \), the maximum horizontal distance is \( 8 \, \text{m} \).