1. Choose the train frame

Since the boy is inside the accelerating train, it is easiest to analyze the motion in the train frame.

• The train has acceleration $a$ forward.
• Therefore, in the train frame, the ball experiences a pseudo-acceleration of magnitude $a$ backward.
• Vertically, the ball has acceleration $g$ downward.

So in the train frame:

• horizontal acceleration $= -a$
• vertical acceleration $= -g$

2. Resolve the initial velocity

The ball is thrown with speed $10\,\text{m/s}$ at angle $60^\circ$.

Hence, $u_x = 10\cos 60^\circ = 5\,\text{m/s}$ $u_y = 10\sin 60^\circ = 5\sqrt{3}\,\text{m/s}$

3. Time to return to the initial height

Vertical motion is unaffected by the train's horizontal acceleration.

For a projectile returning to the same height, $T = \frac{2u_y}{g} = \frac{2(5\sqrt{3})}{g} = \frac{10\sqrt{3}}{g}$

Using $g = 10\,\text{m/s}^2$, $T = \sqrt{3}\,\text{s}$

4. Horizontal displacement in the train frame

In the train frame, horizontal displacement of the ball after time $T$ is $x = u_x T - \frac{1}{2}aT^2$

The boy must move forward by $1.15\,$m to catch it, so the ball is at $x = 1.15\,\text{m}$

Therefore, $1.15 = 5(\sqrt{3}) - \frac{1}{2}a(\sqrt{3})^2$ $1.15 = 5\sqrt{3} - \frac{3a}{2}$

Now use $\sqrt{3} \approx 1.732$: $1.15 = 5(1.732) - 1.5a$ $1.15 = 8.66 - 1.5a$ $1.5a = 8.66 - 1.15 = 7.51$ $a = \frac{7.51}{1.5} \approx 5.01$

Thus, $a \approx 5\,\text{m/s}^2$

5. Final answer

The acceleration of the train is $\boxed{5}$

6. Comparison with stored answer

Stored correct answer: $5$

This matches the derived answer.