A ball of mass 400 g is dropped from a height of 5 m. A boy on the ground hits the ball vertically upwards with a bat with an average force of 100 N, so that it attains a vertical height of 20 m. The time for which the ball remains in contact with the bat is (Take g = 10 m `s^(-2)`)

Options

A. 0.08 s
B. 0.12 s
C. 2 s
D. 5 s

Answer

To solve the problem step by step, we will follow the outlined reasoning and calculations based on the information provided in the question.

### Step 1: Convert mass to kilograms
The mass of the ball is given as 400 grams. To convert this to kilograms:
\[ \text{Mass (m)} = \frac{400 \text{ g}}{1000} = 0.4 \text{ kg} \]

**Hint:** Remember that 1 kg = 1000 g.

### Step 2: Calculate the velocity of the ball just before it hits the bat
The ball is dropped from a height of 5 m. We can use the formula for the velocity of an object falling under gravity:
\[ v = \sqrt{2gh} \]
Where:
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 5 \, \text{m} \)

Substituting the values:
\[ v = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10 \, \text{m/s} \]

**Hint:** The formula \( v = \sqrt{2gh} \) comes from the conservation of energy.

### Step 3: Calculate the initial velocity after being hit by the bat
The ball reaches a height of 20 m after being hit. At the maximum height, the final velocity is 0. We can again use the same formula:
\[ 0 = u^2 + 2(-g)(h) \]
Where:
- \( h = 20 \, \text{m} \)
- \( g = 10 \, \text{m/s}^2 \)

Rearranging gives:
\[ u^2 = 2gh \implies u = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20 \, \text{m/s} \]

**Hint:** The negative sign in the equation accounts for the downward acceleration due to gravity.

### Step 4: Use the change in momentum to find the time of contact
The average force exerted by the bat is given as 100 N. According to Newton's second law:
\[ F = \frac{\Delta p}{\Delta t} \]
Where \( \Delta p \) is the change in momentum.

The initial momentum (just before hitting the bat) is:
\[ p_i = m \cdot (-v) = 0.4 \cdot (-10) = -4 \, \text{kg m/s} \]
The final momentum (after being hit by the bat) is:
\[ p_f = m \cdot u = 0.4 \cdot 20 = 8 \, \text{kg m/s} \]

The change in momentum (\( \Delta p \)):
\[ \Delta p = p_f - p_i = 8 - (-4) = 8 + 4 = 12 \, \text{kg m/s} \]

Now substituting into the force equation:
\[ 100 = \frac{12}{\Delta t} \]
Rearranging gives:
\[ \Delta t = \frac{12}{100} = 0.12 \, \text{s} \]

### Final Answer
The time for which the ball remains in contact with the bat is:
\[ \Delta t = 0.12 \, \text{s} \]

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