A person sitting in the rear end of the compartment throws a ball towards the front end. The ball follows a parabolic path. The train is moving with uniform velocity of `20ms^(-1)`. A person standing outside on the ground also observers the ball. How will the maximum heights `(h_(m))` attained and the ranges (R) seen by thrower and the outside observer compare each other?

Options

A. Same `h_m`, differect R
B. same `h_m`, and R
C. different `h_m`, same R
D. different `h_m`, and R

Explanation

To solve the problem, we need to analyze the motion of the ball as observed by both the person sitting in the train and the observer on the ground.

### Step 1: Understand the Motion of the Ball
The ball is thrown from the rear end of the train towards the front end. The train is moving with a uniform velocity of \(20 \, \text{m/s}\). The ball follows a parabolic path due to the influence of gravity.

### Step 2: Define the Velocity Components
Let’s denote the velocity of the ball with respect to the train as:
- \( v_{x} \) (horizontal component)
- \( v_{y} \) (vertical component)

The velocity of the train is \( 20 \, \text{m/s} \) in the horizontal direction.

### Step 3: Velocity of the Ball with Respect to the Ground
The velocity of the ball with respect to the ground can be expressed as:
- \( v_{x, \text{ground}} = v_{x} + 20 \)
- \( v_{y, \text{ground}} = v_{y} \)

### Step 4: Maximum Height Calculation
The maximum height \( h_m \) attained by the ball depends solely on the vertical component of the initial velocity \( v_{y} \). The formula for maximum height is given by:
\[ h_m = \frac{v_{y}^2}{2g} \]
where \( g \) is the acceleration due to gravity. Since \( v_{y} \) is the same for both observers (the thrower and the observer on the ground), the maximum height \( h_m \) will be the same for both.

### Step 5: Range Calculation
The horizontal range \( R \) of the ball is influenced by the horizontal component of the velocity. The time of flight \( t \) can be determined from the vertical motion:
\[ t = \frac{2v_{y}}{g} \]
The range for the thrower in the train is:
\[ R_{\text{train}} = v_{x} \cdot t \]
For the observer on the ground, the range is:
\[ R_{\text{ground}} = (v_{x} + 20) \cdot t \]
Since the horizontal component of the velocity is different for the two observers, the ranges will differ.

### Conclusion
- The maximum height \( h_m \) attained by the ball is the same for both the thrower and the observer on the ground.
- The range \( R \) seen by the thrower and the observer will be different due to the additional velocity of the train.

### Final Answer
The maximum heights \( h_m \) are the same, but the ranges \( R \) are different.