A ball thrown horizontally with velocity v from the top of a tower of height h reaches the ground in t seconds. If another ball of double the mass is thrown horizontally with velocity 3v from the top of another tower of height 4h it reaches the ground in (seconds). In the above problem if the first ball reaches the ground at a horizontal distance d, the second ball reaches the ground at a horizontal distance

Options

A. 6d
B. 3d
C. 12 d
D. 4d

Answer

To solve the problem step by step, we will analyze the motion of both balls separately and then derive the required distances.

### Step 1: Analyze the first ball
The first ball is thrown horizontally from a height \( h \) with an initial horizontal velocity \( v \). The vertical motion can be described by the equation of motion under gravity:

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

Where:
- \( g \) is the acceleration due to gravity,
- \( t \) is the time taken to reach the ground.

From this equation, we can express \( t \):

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

### Step 2: Calculate the horizontal distance for the first ball
The horizontal distance \( d \) traveled by the first ball can be calculated using:

\[ d = v \cdot t \]

Substituting \( t \) from Step 1:

\[ d = v \cdot \sqrt{\frac{2h}{g}} \]

### Step 3: Analyze the second ball
The second ball is thrown horizontally from a height \( 4h \) with an initial horizontal velocity \( 3v \). The vertical motion is again described by the same equation of motion:

\[ 4h = \frac{1}{2} g t_2^2 \]

From this, we can express \( t_2 \):

\[ t_2 = \sqrt{\frac{8h}{g}} \]

### Step 4: Calculate the horizontal distance for the second ball
The horizontal distance \( d_2 \) traveled by the second ball can be calculated using:

\[ d_2 = 3v \cdot t_2 \]

Substituting \( t_2 \) from Step 3:

\[ d_2 = 3v \cdot \sqrt{\frac{8h}{g}} \]

### Step 5: Relate \( d_2 \) to \( d \)
Now, we can express \( d_2 \) in terms of \( d \):

1. From Step 2, we know \( d = v \cdot \sqrt{\frac{2h}{g}} \).
2. We can express \( \sqrt{\frac{8h}{g}} \) as \( 2\sqrt{\frac{2h}{g}} \).

Thus,

\[ d_2 = 3v \cdot 2\sqrt{\frac{2h}{g}} = 6v \cdot \sqrt{\frac{2h}{g}} = 6d \]

### Final Answer
Therefore, the second ball reaches the ground at a horizontal distance of \( 6d \).

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