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

Explanation

To solve the problem, we need to analyze the motion of the ball thrown by the boy inside the train, taking into account the acceleration of the train and the motion of the ball in both horizontal and vertical directions.

### Step-by-Step Solution:

1. **Identify the Components of the Ball's Velocity:**
The ball is thrown with a speed of \(10 \, \text{m/s}\) at an angle of \(60^\circ\) to the horizontal. We can find the horizontal and vertical components of the velocity:
\[ v_x = v \cos(60^\circ) = 10 \cdot \frac{1}{2} = 5 \, \text{m/s} \]
\[ v_y = v \sin(60^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \]

2. **Calculate the Time of Flight:**
The time of flight of the ball can be calculated using the vertical component of the velocity. The formula for the time of flight when the ball returns to the same height is:
\[ t = \frac{2 v_y}{g} \]
Substituting \(g = 10 \, \text{m/s}^2\):
\[ t = \frac{2 \cdot 5\sqrt{3}}{10} = \sqrt{3} \, \text{s} \]

3. **Determine the Relative Motion:**
During the time \(t\), the boy moves forward by \(1.15 \, \text{m}\) inside the train. The ball also moves forward relative to the train. The acceleration of the train is \(a\), and the ball has no horizontal acceleration relative to the ground.

4. **Set Up the Equation for Displacement:**
The displacement of the ball relative to the train can be expressed as:
\[ \text{Displacement} = \text{Initial Velocity} \cdot t + \frac{1}{2} \cdot \text{Acceleration} \cdot t^2 \]
Here, the initial velocity of the ball relative to the train is \(5 \, \text{m/s}\) and the acceleration of the ball relative to the train is \(-a\):
\[ 1.15 = 5 \cdot \sqrt{3} - \frac{1}{2} a t^2 \]
Substituting \(t = \sqrt{3}\):
\[ 1.15 = 5\sqrt{3} - \frac{1}{2} a \cdot 3 \]
Simplifying this gives:
\[ 1.15 = 5\sqrt{3} - \frac{3a}{2} \]

5. **Solve for Acceleration \(a\):**
Rearranging the equation:
\[ \frac{3a}{2} = 5\sqrt{3} - 1.15 \]
\[ a = \frac{2(5\sqrt{3} - 1.15)}{3} \]
Now, calculating \(5\sqrt{3} \approx 8.66\):
\[ a = \frac{2(8.66 - 1.15)}{3} = \frac{2 \cdot 7.51}{3} \approx 5.01 \, \text{m/s}^2 \]

### Final Answer:
The acceleration of the train is approximately \(5 \, \text{m/s}^2\).