A train is moving along a straight line with a constant acceleration 'a' . A boy standing in the train throws a ball forward with a speed of `10 m//s `, at an angle of `60(@)` to the horizontal. The boy has to move forward by `1.15 m ` inside the train to catch the ball back at the initial height . the acceleration of the train , in `m//s^(2)` , is

Options

A. 3 m `s^(-2)`
B. 5 m `s^(-2)`
C. 8 m `s^(-2)`
D. 6 m `s^(-2)`

Explanation

To solve the problem, we need to analyze the motion of the ball thrown by the boy inside the accelerating train. We will break down the steps to find the acceleration of the train.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Initial speed of the ball, \( v = 10 \, \text{m/s} \)
- Angle of projection, \( \theta = 60^\circ \)
- Distance the boy moves forward in the train, \( s = 1.15 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Time of Flight**:
The time of flight \( t \) for a projectile is given by the formula:
\[ t = \frac{2u \sin \theta}{g} \]
Here, \( u = 10 \, \text{m/s} \) and \( \theta = 60^\circ \).
\[ t = \frac{2 \times 10 \times \sin(60^\circ)}{10} = \frac{20 \times \frac{\sqrt{3}}{2}}{10} = 2\sqrt{3} \, \text{s} \]

3. **Calculate the Horizontal Component of the Velocity**:
The horizontal component of the initial velocity \( v_x \) is:
\[ v_x = v \cos \theta = 10 \cos(60^\circ) = 10 \times \frac{1}{2} = 5 \, \text{m/s} \]

4. **Set Up the Equation of Motion**:
The boy moves forward with respect to the train while the ball is in the air. The relative motion can be described using the second equation of motion:
\[ s = v_{bt} t + \frac{1}{2} a_{bt} t^2 \]
Where:
- \( s = 1.15 \, \text{m} \) (distance moved by the boy)
- \( v_{bt} = v_x - a \cdot t \) (the effective horizontal velocity of the boy with respect to the train)
- \( a_{bt} = -a \) (the acceleration of the ball with respect to the train)

5. **Substituting Values**:
Substitute \( v_{bt} \) and \( t \) into the equation:
\[ 1.15 = (5 - a) \cdot 2\sqrt{3} + \frac{1}{2} (-a) (2\sqrt{3})^2 \]
Simplifying the equation:
\[ 1.15 = (5 - a) \cdot 2\sqrt{3} - 2a \cdot 3 \]
\[ 1.15 = 10\sqrt{3} - 2a\sqrt{3} - 6a \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ 6a + 2a\sqrt{3} = 10\sqrt{3} - 1.15 \]
\[ a(6 + 2\sqrt{3}) = 10\sqrt{3} - 1.15 \]
\[ a = \frac{10\sqrt{3} - 1.15}{6 + 2\sqrt{3}} \]

7. **Calculating the Value of \( a \)**:
Now, we can calculate the value of \( a \):
\[ a \approx \frac{10 \times 1.732 - 1.15}{6 + 2 \times 1.732} \]
\[ a \approx \frac{17.32 - 1.15}{6 + 3.464} \approx \frac{16.17}{9.464} \approx 1.71 \, \text{m/s}^2 \]

### Final Result:
The acceleration of the train is approximately \( 1.71 \, \text{m/s}^2 \).