A ball falls freely from a height of 45m. When the ball is at a height of 25m, it explodes into two equal piece. One of them moves horizontally with a speed of `10ms^(-1)`. The distance between the two pieces on the ground is

Options

A. 20 m
B. 30 m
C. 40 m
D. 60 m

Explanation

To solve the problem step by step, we will follow these procedures:

### Step 1: Determine the height from which the ball falls
The ball falls freely from a height of 45 m and explodes at a height of 25 m. Therefore, the distance fallen before the explosion is:
\[ h = 45 \, \text{m} - 25 \, \text{m} = 20 \, \text{m} \]

### Step 2: Calculate the velocity of the ball just before the explosion
Using the equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( u = 0 \) (initial velocity)
- \( a = g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( s = 20 \, \text{m} \) (distance fallen)

Substituting the values:
\[ v^2 = 0 + 2 \times 10 \times 20 \]
\[ v^2 = 400 \]
\[ v = \sqrt{400} = 20 \, \text{m/s} \]

### Step 3: Analyze the explosion and the motion of the pieces
When the ball explodes into two equal pieces, one piece moves horizontally with a speed of \( 10 \, \text{m/s} \). Let’s denote the mass of each piece as \( \frac{M}{2} \).

By conservation of momentum in the horizontal direction:
\[ 0 = \left(\frac{M}{2}\right)(10) + \left(\frac{M}{2}\right)(v_x) \]
where \( v_x \) is the horizontal velocity of the other piece. Rearranging gives:
\[ \frac{M}{2} v_x = -\frac{M}{2} \cdot 10 \]
\[ v_x = -10 \, \text{m/s} \]
This means the second piece moves horizontally in the opposite direction at \( 10 \, \text{m/s} \).

### Step 4: Calculate the time taken for both pieces to reach the ground
Both pieces fall from a height of 25 m. The time \( t \) taken to fall can be calculated using:
\[ s = ut + \frac{1}{2}gt^2 \]
Here, \( s = 25 \, \text{m} \), \( u = 0 \), and \( g = 10 \, \text{m/s}^2 \):
\[ 25 = 0 + \frac{1}{2} \cdot 10 \cdot t^2 \]
\[ 25 = 5t^2 \]
\[ t^2 = 5 \]
\[ t = \sqrt{5} \approx 2.24 \, \text{s} \]

### Step 5: Calculate the horizontal distance traveled by each piece
- For the piece moving horizontally at \( 10 \, \text{m/s} \):
\[ \text{Distance}_1 = 10 \, \text{m/s} \times t = 10 \times \sqrt{5} \approx 22.36 \, \text{m} \]

- For the piece moving in the opposite direction at \( -10 \, \text{m/s} \):
\[ \text{Distance}_2 = -10 \, \text{m/s} \times t = -10 \times \sqrt{5} \approx -22.36 \, \text{m} \]

### Step 6: Calculate the total distance between the two pieces on the ground
The total distance between the two pieces is the sum of the absolute distances:
\[ \text{Distance between pieces} = |22.36| + |22.36| = 22.36 + 22.36 = 44.72 \, \text{m} \]

### Final Answer
The distance between the two pieces on the ground is approximately \( 44.72 \, \text{m} \).

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