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 constat 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 determine the constant force \( F \) applied to the cricket ball by the bowling machine. We can do this by equating the work done by the force to the change in potential energy of the ball when it reaches its maximum height.

### Step-by-Step Solution:

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

2. **Calculate the potential energy at maximum height:**
The potential energy (PE) gained by the ball when it reaches the maximum height is given by:
\[ PE = mgh \]
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 the force:**
The work done \( W \) by the force \( F \) while the machine moves a distance \( d \) is given by:
\[ W = F \cdot d \]
Since the force is applied in the direction of the displacement, we can write:
\[ W = F \times 0.2 \, \text{m} \]

4. **Set the work done equal to the potential energy:**
The work done on the ball is equal to the potential energy gained:
\[ F \cdot 0.2 = 30 \]

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

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

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