A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine so that it rises to a maximum height of 20 m after leaving the machine. If the part pushing the ball applies a constant force F on the ball and moves horizontally a distance of 0.2 m while launching the ball, the value of F (in N) is `(g = 10 ms^(-2))` __________.

Explanation

To solve the problem, we need to find the constant force \( F \) applied to the cricket ball by the bowling machine while it moves horizontally a distance of 0.2 m. The ball reaches a maximum height of 20 m after being launched.

### Step-by-step Solution:

1. **Identify the given values:**
- Mass of the cricket ball, \( m = 0.15 \, \text{kg} \)
- Maximum height reached, \( h = 20 \, \text{m} \)
- Distance moved by the machine while launching the ball, \( d = 0.2 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the potential energy (PE) at maximum height:**
The potential energy at the maximum height can be calculated using the formula:
\[ PE = mgh \]
Substituting the values:
\[ PE = 0.15 \, \text{kg} \times 10 \, \text{m/s}^2 \times 20 \, \text{m} = 30 \, \text{J} \]

3. **Relate work done to potential energy:**
The work done by the force \( F \) while moving the ball a distance of 0.2 m is given by:
\[ W = F \times d \]
Since the work done on the ball is converted into potential energy at the maximum height, we can set the work done equal to the potential energy:
\[ F \times 0.2 = 30 \]

4. **Solve for the force \( F \):**
Rearranging the equation to find \( F \):
\[ F = \frac{30}{0.2} = 150 \, \text{N} \]

### Final Answer:
The value of the force \( F \) is \( 150 \, \text{N} \).

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