Two small balls having the same mass and charge and located on the same vertical at heights `h_(1)` and `h_(2)` are thrown in the same direction along the horizontal at the same velocity v. The first vall touches the ground at a horizontal distance R from the initial vertical position. At waht heigth `h_(2)` will the second ball be at this instant? Neglect any frictional resistance of air and the effect of any induced charge on the ground.

Explanation

To solve the problem, we need to analyze the motion of the two balls and their respective heights when the first ball touches the ground. Here’s the step-by-step solution:

### Step 1: Understand the Motion of the First Ball
The first ball is thrown horizontally from a height \( h_1 \) and travels a horizontal distance \( R \) before hitting the ground. The time \( t \) it takes to fall from height \( h_1 \) can be calculated using the equation of motion under gravity:

\[ h_1 = \frac{1}{2} g t^2 \]

From this, we can solve for \( t \):

\[ t = \sqrt{\frac{2h_1}{g}} \]

### Step 2: Calculate the Horizontal Distance
The horizontal distance \( R \) covered by the first ball can be expressed in terms of its initial velocity \( v \):

\[ R = v \cdot t \]

Substituting the expression for \( t \):

\[ R = v \cdot \sqrt{\frac{2h_1}{g}} \]

### Step 3: Analyze the Second Ball's Motion
The second ball is thrown from a height \( h_2 \) at the same time and with the same horizontal velocity \( v \). It will also take the same time \( t \) to reach the ground as the first ball, since they are thrown simultaneously.

### Step 4: Calculate the Height of the Second Ball
We know that the second ball will fall for the same time \( t \):

\[ h_2 = \frac{1}{2} g t^2 \]

Substituting \( t \) from Step 1 into this equation:

\[ h_2 = \frac{1}{2} g \left(\sqrt{\frac{2h_1}{g}}\right)^2 \]

This simplifies to:

\[ h_2 = h_1 \]

### Step 5: Determine the Height of the Second Ball at Distance \( R \)
Now, we need to find the height of the second ball when the first ball touches the ground at distance \( R \). Since both balls are thrown at the same time and with the same velocity, the second ball will also have fallen a distance equal to \( h_2 \) after time \( t \).

The height of the second ball when the first ball touches the ground can be expressed as:

\[ h_2 = h_1 - \frac{1}{2} g t^2 \]

Substituting \( t^2 \) from the first step:

\[ h_2 = h_1 - h_1 = 0 \]

### Conclusion
Thus, at the instant the first ball touches the ground, the height \( h_2 \) of the second ball will be \( 0 \) if both balls are thrown from the same height.