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 step by step, we will analyze the motion of the ball and the train separately, and then relate them to find the acceleration of the train.

### Step 1: Understand the problem
The train is accelerating, and a body inside the train throws a ball at a speed of \(10 \, \text{m/s}\) at an angle of \(60^\circ\) to the horizontal. The body has to move forward by \(1.15 \, \text{m}\) to catch the ball back at the initial height.

### Step 2: Break down the initial velocity of the ball
The initial velocity of the ball can be broken down into horizontal and vertical components:
- Horizontal component \(u_x = 10 \cos(60^\circ) = 10 \times \frac{1}{2} = 5 \, \text{m/s}\)
- Vertical component \(u_y = 10 \sin(60^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s}\)

### Step 3: Calculate the time of flight for the ball
The vertical motion of the ball is influenced by gravity. The time of flight \(t\) can be calculated using the equation for vertical motion:
\[ y = u_y t - \frac{1}{2} g t^2 \]
Since the ball returns to the same height, \(y = 0\):
\[ 0 = 5\sqrt{3} t - \frac{1}{2} g t^2 \]
Factoring out \(t\):
\[ t(5\sqrt{3} - \frac{1}{2} g t) = 0 \]
This gives us \(t = 0\) or \(t = \frac{10\sqrt{3}}{g}\). Using \(g = 10 \, \text{m/s}^2\) for simplicity, we find:
\[ t = \frac{10\sqrt{3}}{10} = \sqrt{3} \, \text{s} \]

### Step 4: Analyze horizontal motion
The horizontal distance traveled by the ball in time \(t\) is given by:
\[ s = u_x t \]
The body in the train also moves forward with the train's acceleration \(a\). The total distance covered by the body in the train is:
\[ s = u_x t + \frac{1}{2} a t^2 \]
Substituting \(u_x = 5 \, \text{m/s}\) and \(t = \sqrt{3} \, \text{s}\):
\[ 1.15 = 5 \cdot \sqrt{3} + \frac{1}{2} a (\sqrt{3})^2 \]
This simplifies to:
\[ 1.15 = 5\sqrt{3} + \frac{3}{2} a \]

### Step 5: Solve for acceleration \(a\)
Rearranging the equation:
\[ \frac{3}{2} a = 1.15 - 5\sqrt{3} \]
\[ a = \frac{2(1.15 - 5\sqrt{3})}{3} \]

### Step 6: Substitute numerical values
Calculating \(5\sqrt{3}\):
\[ 5\sqrt{3} \approx 5 \times 1.732 = 8.66 \]
Now substituting:
\[ a = \frac{2(1.15 - 8.66)}{3} = \frac{2(-7.51)}{3} \approx -5.007 \, \text{m/s}^2 \]
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 \(5 \, \text{m/s}^2\).