A ball is thrown from a comer of room of length 20 m and height 5 m such that the ball moves along the length of room just touching the ceiling and drops on the other corner. Find out
(i) Angle of projection
(ii) Velocity of projection
(iii) Time of flight

Answer

To solve the problem, we need to find the angle of projection, velocity of projection, and time of flight for a ball thrown from one corner of a room to the opposite corner, touching the ceiling. The room has a length of 20 m and a height of 5 m.

### Step-by-Step Solution:

#### Given:
- Length of the room (Range, R) = 20 m
- Height of the room (Maximum height, H) = 5 m
- Acceleration due to gravity (g) = 10 m/s² (assuming standard value)

### (i) Finding the Angle of Projection (θ)

1. **Use the relationship between range and height:**
The relationship can be derived from the equations of projectile motion:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

2. **Rearranging the equations:**
From the range equation:
\[ R = \frac{u^2 \sin 2\theta}{g} \implies u^2 = \frac{Rg}{\sin 2\theta} \]
From the height equation:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \implies u^2 = \frac{2Hg}{\sin^2 \theta} \]

3. **Setting the two expressions for \( u^2 \) equal:**
\[ \frac{Rg}{\sin 2\theta} = \frac{2Hg}{\sin^2 \theta} \]

4. **Substituting values:**
\[ \frac{20 \cdot 10}{\sin 2\theta} = \frac{2 \cdot 5 \cdot 10}{\sin^2 \theta} \]
Simplifying gives:
\[ \frac{200}{\sin 2\theta} = \frac{100}{\sin^2 \theta} \]

5. **Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):**
\[ \frac{200}{2 \sin \theta \cos \theta} = \frac{100}{\sin^2 \theta} \]
Cross-multiplying leads to:
\[ 200 \sin^2 \theta = 200 \sin \theta \cos \theta \]
Dividing both sides by 100:
\[ 2 \sin^2 \theta = 2 \sin \theta \cos \theta \]
Simplifying gives:
\[ \sin \theta = \cos \theta \]
Thus, \( \theta = 45^\circ \).

### (ii) Finding the Velocity of Projection (u)

1. **Using the range formula:**
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Substituting \( \theta = 45^\circ \):
\[ R = \frac{u^2 \cdot 1}{10} \quad (\text{since } \sin 90^\circ = 1) \]
\[ 20 = \frac{u^2}{10} \implies u^2 = 200 \implies u = \sqrt{200} = 10\sqrt{2} \, \text{m/s} \]

### (iii) Finding the Time of Flight (T)

1. **Using the time of flight formula:**
\[ T = \frac{2u \sin \theta}{g} \]
Substituting \( u = 10\sqrt{2} \) and \( \theta = 45^\circ \):
\[ T = \frac{2 \cdot 10\sqrt{2} \cdot \frac{1}{\sqrt{2}}}{10} \]
Simplifying gives:
\[ T = \frac{20}{10} = 2 \, \text{s} \]

### Final Answers:
- (i) Angle of Projection, \( \theta = 45^\circ \)
- (ii) Velocity of Projection, \( u = 10\sqrt{2} \, \text{m/s} \)
- (iii) Time of Flight, \( T = 2 \, \text{s} \)