A ball is dropped from a height of 80 m on a floor. At each collision, the ball loses half of its speed. Plot the speed-time graph and velocity-time graph of its motion till two collisions With the floor. [Take `g=10 m//s^2`]

Explanation

To solve the problem of a ball dropped from a height of 80 meters and losing half of its speed after each collision, we will follow these steps:

### Step 1: Calculate the speed just before the first collision
We can use the kinematic equation:
\[ V^2 = U^2 + 2gH \]
Where:
- \( V \) = final velocity
- \( U \) = initial velocity (0 m/s, since the ball is dropped)
- \( g \) = acceleration due to gravity (10 m/s²)
- \( H \) = height (80 m)

Substituting the values:
\[ V^2 = 0 + 2 \times 10 \times 80 \]
\[ V^2 = 1600 \]
\[ V = \sqrt{1600} = 40 \text{ m/s} \]

### Step 2: Calculate the time taken to reach the ground
Using the equation:
\[ V = U + gt \]
Substituting the known values:
\[ 40 = 0 + 10t \]
\[ t = \frac{40}{10} = 4 \text{ seconds} \]

### Step 3: Speed after the first collision
After the first collision, the ball loses half of its speed:
\[ \text{New speed} = \frac{40}{2} = 20 \text{ m/s} \]

### Step 4: Calculate the time taken to reach the maximum height after the first collision
Using the same formula:
\[ V = U - gt \]
Where \( V = 0 \) (at the maximum height) and \( U = 20 \):
\[ 0 = 20 - 10t \]
\[ 10t = 20 \]
\[ t = 2 \text{ seconds} \]

### Step 5: Calculate the height reached after the first collision
Using the formula:
\[ H = Ut - \frac{1}{2}gt^2 \]
Substituting the values:
\[ H = 20 \times 2 - \frac{1}{2} \times 10 \times (2^2) \]
\[ H = 40 - 20 = 20 \text{ meters} \]

### Step 6: Speed just before the second collision
The ball falls from a height of 20 m, so we calculate the speed just before the second collision:
\[ V^2 = U^2 + 2gH \]
Where \( U = 0 \):
\[ V^2 = 0 + 2 \times 10 \times 20 \]
\[ V^2 = 400 \]
\[ V = \sqrt{400} = 20 \text{ m/s} \]

### Step 7: Time taken to reach the ground for the second collision
Using:
\[ V = U + gt \]
Where \( U = 0 \):
\[ 20 = 0 + 10t \]
\[ t = \frac{20}{10} = 2 \text{ seconds} \]

### Step 8: Speed after the second collision
After the second collision, the ball again loses half of its speed:
\[ \text{New speed} = \frac{20}{2} = 10 \text{ m/s} \]

### Step 9: Calculate the time taken to reach the maximum height after the second collision
Using:
\[ V = U - gt \]
Where \( U = 10 \):
\[ 0 = 10 - 10t \]
\[ 10t = 10 \]
\[ t = 1 \text{ second} \]

### Step 10: Speed-time and velocity-time graphs
1. **Speed-time graph**:
- From 0 to 4 seconds, speed increases from 0 to 40 m/s.
- At 4 seconds, speed drops to 20 m/s (first collision).
- From 4 to 6 seconds, speed decreases from 20 to 0 m/s.
- At 6 seconds, speed increases from 0 to 20 m/s (second collision).
- At 8 seconds, speed drops to 10 m/s (second collision).
- From 8 to 9 seconds, speed decreases from 10 to 0 m/s.

2. **Velocity-time graph**:
- From 0 to 4 seconds, velocity increases from 0 to -40 m/s (downward).
- At 4 seconds, velocity changes to -20 m/s (upward after first collision).
- From 4 to 6 seconds, velocity decreases from -20 to 0 m/s.
- From 6 to 8 seconds, velocity increases from 0 to -20 m/s (downward).
- At 8 seconds, velocity changes to -10 m/s (upward after second collision).
- From 8 to 9 seconds, velocity decreases from -10 to 0 m/s.