A `15 gm` ball is shot from a spring whose spring has a force constant of `600 N//m`. The spring is compressed by `5 cm`. The greatest possible horizontal range of the ball for this compression is

Options

A. 6 m
B. 8 m
C. 10 m
D. 12 m

Explanation

To find the greatest possible horizontal range of a ball shot from a spring, we can follow these steps:

### Step 1: Identify the given values
- Mass of the ball \( m = 15 \, \text{g} = 15 \times 10^{-3} \, \text{kg} \)
- Spring constant \( k = 600 \, \text{N/m} \)
- Compression of the spring \( x = 5 \, \text{cm} = 0.05 \, \text{m} \)

### Step 2: Calculate the potential energy stored in the spring
The potential energy \( PE \) stored in a compressed spring is given by the formula:
\[ PE = \frac{1}{2} k x^2 \]
Substituting the values:
\[ PE = \frac{1}{2} \times 600 \, \text{N/m} \times (0.05 \, \text{m})^2 \]
Calculating \( (0.05)^2 = 0.0025 \):
\[ PE = \frac{1}{2} \times 600 \times 0.0025 = 0.75 \, \text{J} \]

### Step 3: Relate potential energy to kinetic energy
When the spring releases, the potential energy is converted into kinetic energy \( KE \):
\[ KE = \frac{1}{2} m u^2 \]
Setting the potential energy equal to the kinetic energy:
\[ 0.75 = \frac{1}{2} \times (15 \times 10^{-3}) \times u^2 \]

### Step 4: Solve for the initial velocity \( u \)
Rearranging the equation:
\[ u^2 = \frac{0.75 \times 2}{15 \times 10^{-3}} = \frac{1.5}{15 \times 10^{-3}} = \frac{1.5}{0.015} = 100 \]
Thus,
\[ u = \sqrt{100} = 10 \, \text{m/s} \]

### Step 5: Calculate the maximum horizontal range
The maximum range \( R_{max} \) for projectile motion is given by:
\[ R_{max} = \frac{u^2}{g} \]
Where \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity). Substituting the value of \( u \):
\[ R_{max} = \frac{10^2}{10} = \frac{100}{10} = 10 \, \text{m} \]

### Final Answer
The greatest possible horizontal range of the ball is \( \boxed{10 \, \text{m}} \).