A train is moving along a straight line with a constant acceleration a. A body standing in the train throws a ball forward with a speed of `10ms^(-1)`, at an angle of `60^(@)` to the horizontal . The body has to move forward by 1.15 m inside the train to cathc the ball back to the initial height. the acceleration of the train. in `ms^(-2)` , is:

Explanation

To solve the problem, we need to determine the acceleration of the train given the conditions of the ball being thrown and the distance the body moves inside the train.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Initial speed of the ball, \( u = 10 \, \text{m/s} \)
- Angle of projection, \( \theta = 60^\circ \)
- Distance moved by the body in the train, \( x = 1.15 \, \text{m} \)
- 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:
\[ T = \frac{u \sin \theta}{g} \]
Substituting the values:
\[ T = \frac{10 \sin 60^\circ}{10} = \sin 60^\circ = \frac{\sqrt{3}}{2} \]
Therefore, the time of flight \( T = \sqrt{3} \, \text{s} \).

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

4. **Set Up the Equation for Distance**:
The distance covered by the body in the train can be expressed as:
\[ x = u_x T + \frac{1}{2} a T^2 \]
Substituting the known values:
\[ 1.15 = 5 \cdot \sqrt{3} + \frac{1}{2} a (\sqrt{3})^2 \]
Simplifying:
\[ 1.15 = 5 \sqrt{3} + \frac{3}{2} a \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{3}{2} a = 1.15 - 5 \sqrt{3} \]

6. **Calculate \( 5 \sqrt{3} \)**:
The value of \( \sqrt{3} \approx 1.732 \):
\[ 5 \sqrt{3} \approx 5 \times 1.732 \approx 8.66 \]
Now substituting back:
\[ \frac{3}{2} a = 1.15 - 8.66 = -7.51 \]

7. **Solve for \( a \)**:
\[ a = \frac{-7.51 \times 2}{3} \approx -5.007 \, \text{m/s}^2 \]
Since acceleration cannot be negative in this context, we take the absolute value:
\[ a \approx 5 \, \text{m/s}^2 \]

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

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