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 degrees 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

Step 1: Find the time of flight of the ball.

The vertical motion of the ball is independent of the train's horizontal acceleration.

$$u_y = u \sin 60^\circ = 10 \sin 60^\circ = 5\sqrt{3} \text{ m/s}$$ $$T = \frac{2u_y}{g}$$

Using $g = 10 \text{ m/s}^2$:

$$T = \frac{2 \times 5\sqrt{3}}{10} = \sqrt{3} \text{ s}$$

Step 2: Analyze the horizontal motion relative to the train.

The initial horizontal velocity of the ball relative to the train is:

$$u_x = u \cos 60^\circ = 10 \cos 60^\circ = 5 \text{ m/s}$$

Since the train accelerates forward at $a$, the ball has a relative backward acceleration of $a$ inside the train.

$$a_{\text{rel}} = -a$$

Step 3: Set up the relative displacement equation.

The displacement of the ball relative to the train during the time of flight $T$ must equal the distance the boy moves forward to catch it.

$$x_{\text{rel}} = u_x T - \frac{1}{2} a T^2$$ $$5\sqrt{3} - \frac{1}{2} a (\sqrt{3})^2 = 1.15$$

Step 4: Solve for the acceleration $a$.

Substitute the value of $\sqrt{3} \approx 1.732$ and solve.

$$5(1.732) - 1.5a = 1.15$$ $$8.66 - 1.5a = 1.15$$ $$1.5a = 7.51$$ $$a = \frac{7.51}{1.5} \approx 5 \text{ m/s}^2$$

Answer: The acceleration of the train is $5\text{ m/s}^{2}$.