A ball is thrown from the ground to clear a wall 3 m high at a distance of 6 m and falls 18 m away from the wall. Find the angle of projection of ball.

Explanation

To find the angle of projection of the ball, we can use the equation of trajectory for projectile motion. Here’s the step-by-step solution:

### Step 1: Understand the problem
The ball is thrown from the ground and must clear a wall that is 3 meters high, located 6 meters away from the starting point. The ball lands 18 meters beyond the wall.

### Step 2: Determine the horizontal range (R)
The total horizontal range (R) of the projectile can be calculated by adding the distance to the wall and the distance from the wall to where the ball lands:
\[ R = 6 \, \text{m} + 18 \, \text{m} = 24 \, \text{m} \]

### Step 3: Identify the coordinates at the wall
At the wall, the horizontal distance (x) is 6 meters and the vertical height (y) is 3 meters. Thus, we have:
- \( x = 6 \, \text{m} \)
- \( y = 3 \, \text{m} \)

### Step 4: Use the equation of trajectory
The equation of trajectory for a projectile is given by:
\[ y = x \tan \theta \left(1 - \frac{x}{R}\right) \]
Substituting the known values into the equation:
\[ 3 = 6 \tan \theta \left(1 - \frac{6}{24}\right) \]

### Step 5: Simplify the equation
First, simplify the term \( 1 - \frac{6}{24} \):
\[ 1 - \frac{6}{24} = 1 - \frac{1}{4} = \frac{3}{4} \]
Now substitute this back into the equation:
\[ 3 = 6 \tan \theta \cdot \frac{3}{4} \]

### Step 6: Solve for \( \tan \theta \)
Rearranging the equation gives:
\[ 3 = \frac{18}{4} \tan \theta \]
\[ 3 = \frac{9}{2} \tan \theta \]
Now, isolate \( \tan \theta \):
\[ \tan \theta = \frac{3 \cdot 2}{9} = \frac{6}{9} = \frac{2}{3} \]

### Step 7: Find the angle \( \theta \)
To find the angle of projection \( \theta \), we take the arctangent:
\[ \theta = \tan^{-1}\left(\frac{2}{3}\right) \]

### Final Answer
The angle of projection of the ball is:
\[ \theta = \tan^{-1}\left(\frac{2}{3}\right) \]

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