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.

Options

A. `tan^(-1) ((3)/(2))`
B. `tan^(-1)((2)/(3))`
C. `tan^(-1)((1)/(2))`
D. `tan^(-1)((3)/(4))`

Explanation

To solve the problem of finding the angle of projection of a ball thrown to clear a wall, we can follow these steps:

### Step 1: Understand the Problem
We have a ball thrown from the ground that needs to clear a wall that is 3 meters high and located 6 meters away from the throw point. The ball lands 18 meters beyond the wall, making the total horizontal distance from the throw point to where the ball lands 24 meters.

### Step 2: Set Up the Variables
- Height of the wall (h) = 3 m
- Distance to the wall (d) = 6 m
- Total distance (R) = 24 m (6 m to the wall + 18 m beyond the wall)
- Acceleration due to gravity (g) = 9.81 m/s² (approximately)

### Step 3: Use the Range Formula
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Where:
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

From the problem, we have:
\[ 24 = \frac{u^2 \sin 2\theta}{g} \]
Rearranging gives:
\[ \frac{u^2}{g} = \frac{24}{\sin 2\theta} \]
Let’s call this Equation (1).

### Step 4: Use the Trajectory Equation
The height \( y \) of the projectile at a horizontal distance \( x \) is given by:
\[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \]
Substituting \( y = 3 \) m and \( x = 6 \) m into the equation:
\[ 3 = 6 \tan \theta - \frac{g \cdot 6^2}{2 u^2 \cos^2 \theta} \]
This simplifies to:
\[ 3 = 6 \tan \theta - \frac{36g}{2u^2 \cos^2 \theta} \]
\[ 3 = 6 \tan \theta - \frac{18g}{u^2 \cos^2 \theta} \]
Now, substituting \( \frac{u^2}{g} \) from Equation (1):
\[ 3 = 6 \tan \theta - \frac{18}{\frac{24}{\sin 2\theta} \cos^2 \theta} \]
This simplifies to:
\[ 3 = 6 \tan \theta - \frac{18 \sin 2\theta \cos^2 \theta}{24} \]
\[ 3 = 6 \tan \theta - \frac{3 \sin 2\theta \cos^2 \theta}{4} \]

### Step 5: Solve for \( \tan \theta \)
Rearranging gives:
\[ 3 + \frac{3 \sin 2\theta \cos^2 \theta}{4} = 6 \tan \theta \]
From the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ 3 + \frac{3 \cdot 2 \sin \theta \cos^3 \theta}{4} = 6 \tan \theta \]
This leads us to find \( \tan \theta \) in terms of \( \sin \theta \) and \( \cos \theta \).

### Step 6: Final Calculation
After simplifying, we find:
\[ \tan \theta = \frac{6}{9} = \frac{2}{3} \]
Thus, the angle of projection \( \theta \) is:
\[ \theta = \tan^{-1}\left(\frac{2}{3}\right) \]

### Conclusion
The angle of projection of the ball is \( \tan^{-1}\left(\frac{2}{3}\right) \).

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