Animated Solution for Physics - Kinematics: 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 $10\text{ m/s}$, at an angle of $60^{\circ }$ to the horizontal. The boy has to move forward by $1.15\text{ m}$ inside the train to catch the ball back at the initial height. The acceleration of the train, in ${\text{m/s}}^2$, is.

Imagine you are standing in a train that is smoothly accelerating out of the station. You toss a ball forward. To your surprise, the ball doesn't just follow a simple parabola relative to you—it seems to drift backward as the train speeds up beneath it! This problem captures that exact thrilling scenario.

We can solve this by jumping out of the train—conceptually, of course! By analyzing the motion from the stationary ground frame, the complex non-inertial pseudo-forces disappear, leaving us with pure, elegant kinematics. Let's assume the train starts from rest. If it had an initial velocity, it would simply add the exact same horizontal displacement to both the ball and the train, perfectly canceling out in our final equation.

First, let's look at the vertical motion. Gravity is the only force acting vertically. The train's horizontal acceleration has absolutely no jurisdiction over the vertical realm. This independence of perpendicular vectors is one of the most beautiful symmetries in physics.

Because the vertical motion is unaffected, we can calculate the time of flight $T$ using our standard projectile formula:

Substituting the given values of $u=10\text{ m/s}$ and $\theta =60^{\circ }$, we get:

Now, where does the ball land in the ground frame? Since there is no horizontal force acting on the ball once it leaves the boy's hand, it travels with a constant horizontal velocity $u_x=u\cos 60^{\circ }$. Its total horizontal displacement is simply velocity multiplied by time:

Meanwhile, what is the boy doing? He is surfing the accelerating train! The train itself moves a distance of $\frac{1}{2}a{T}^2$. But the boy isn't glued to the floor; he actively runs forward by $1.15\text{ m}$ inside the train to catch the ball. Therefore, his total absolute displacement relative to the ground is:

The Catch

For the boy to successfully catch the ball, their absolute horizontal displacements from the starting point must be exactly equal. We set up our master equation by equating $S_{ball}$ and $S_{boy}$:

Squaring the $\sqrt{3}$ gives us $3$, turning the equation into a simple linear relation:

Using the standard approximation $\sqrt{3}\approx 1.732$, we find that $5\times 1.732=8.66$. Substituting this back in:

Rounding to the nearest integer, we find the acceleration of the train is exactly $5{\text{ m/s}}^2$. By shifting our perspective to the ground frame, a complex non-inertial problem transformed into a beautiful, straightforward kinematic race!