In the above problem, what is the angle of projection with horizontal ?

Options

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

Explanation

To find the angle of projection with the horizontal, we can follow these steps:

### Step 1: Understand the Problem
We are given two velocities: the velocity of the ball with respect to the ground (let's denote it as \( V_{BG} \)) and the velocity of the truck with respect to the ground (denote it as \( V_{TG} \)). We need to find the angle of projection \( \theta \) with respect to the horizontal.

### Step 2: Identify the Given Values
From the problem, we know:
- The velocity of the ball with respect to the ground, \( V_{BG} = 20 \, \text{m/s} \)
- The velocity of the truck with respect to the ground, \( V_{TG} = 15 \, \text{m/s} \)

### Step 3: Use the Tangent Function
The tangent of the angle \( \theta \) can be expressed as the ratio of the vertical component of the ball's velocity to the horizontal component (which is the velocity of the truck):
\[ \tan \theta = \frac{V_{BG}}{V_{TG}} \]

### Step 4: Substitute the Values
Now, substitute the values into the equation:
\[ \tan \theta = \frac{20 \, \text{m/s}}{15 \, \text{m/s}} = \frac{20}{15} = \frac{4}{3} \]

### Step 5: Calculate the Angle
To find the angle \( \theta \), we take the arctangent (inverse tangent) of \( \frac{4}{3} \):
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 6: Conclusion
Thus, the angle of projection with respect to the horizontal is:
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]