A ball of mass 1 kg is dropped from height 9.8 m, strikes with ground and rebounds at height 4.9 m, if the time of contact between ball and ground is 0.1 s then find impulse and average force acting on ball.

Options

A. 23.52 N-s,235.2 N
B. 235.2 N-s,23.53 N
C. 42.5 N-s,525 N
D. 52.5 N-s,525N

Explanation

To solve the problem step-by-step, we will calculate the impulse and average force acting on the ball when it strikes the ground and rebounds.

### Step 1: Calculate the velocity just before hitting the ground (V1)
The ball is dropped from a height of 9.8 m. We can use the equation of motion to find the velocity just before it hits the ground.

Using the equation:
\[ V^2 = U^2 + 2gh \]

Where:
- \( V \) = final velocity (just before hitting the ground)
- \( U \) = initial velocity (0 m/s, since the ball is dropped)
- \( g \) = acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \))
- \( h \) = height (9.8 m)

Substituting the values:
\[ V_1^2 = 0 + 2 \times 9.8 \times 9.8 \]
\[ V_1^2 = 2 \times 9.8 \times 9.8 \]
\[ V_1^2 = 192.08 \]
\[ V_1 = \sqrt{192.08} \]
\[ V_1 \approx 13.86 \, \text{m/s} \]

### Step 2: Calculate the velocity just after rebounding (U2)
The ball rebounds to a height of 4.9 m. We can again use the equation of motion to find the velocity just after it rebounds.

Using the same equation:
\[ U^2 = V^2 + 2gh \]

Where:
- \( U \) = initial velocity after rebounding (which we need to find)
- \( V \) = final velocity at the peak of the rebound (0 m/s)
- \( g \) = \( 9.8 \, \text{m/s}^2 \)
- \( h \) = height (4.9 m)

Substituting the values:
\[ 0 = U_2^2 - 2 \times 9.8 \times 4.9 \]
\[ U_2^2 = 2 \times 9.8 \times 4.9 \]
\[ U_2^2 = 96.08 \]
\[ U_2 = \sqrt{96.08} \]
\[ U_2 \approx 9.80 \, \text{m/s} \]

### Step 3: Calculate the change in velocity (ΔV)
The change in velocity (ΔV) is given by:
\[ \Delta V = U_2 - (-V_1) \]
\[ \Delta V = 9.80 - (-13.86) \]
\[ \Delta V = 9.80 + 13.86 \]
\[ \Delta V = 23.66 \, \text{m/s} \]

### Step 4: Calculate the impulse (J)
Impulse is defined as the product of mass and change in velocity:
\[ J = m \cdot \Delta V \]
Where:
- \( m = 1 \, \text{kg} \)

Substituting the values:
\[ J = 1 \cdot 23.66 \]
\[ J = 23.66 \, \text{kg m/s} \]

### Step 5: Calculate the average force (F)
The average force can be calculated using the formula:
\[ F = \frac{J}{t} \]
Where:
- \( t = 0.1 \, \text{s} \)

Substituting the values:
\[ F = \frac{23.66}{0.1} \]
\[ F = 236.6 \, \text{N} \]

### Final Answers
- Impulse (J) = 23.66 kg m/s
- Average Force (F) = 236.6 N