To solve the problem of a ball projected vertically upward with an initial speed of 20 m/s and gravitational acceleration \( g = 10 \, \text{m/s}^2 \), we will go through the following steps:
### Step 1: Identify Given Data
- Initial speed, \( u = 20 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (acting downward, hence we will use \( -g \) in our equations)
### Step 2: Calculate the Distance Covered in the Second Second
To find the distance covered in the second second of motion, we can use the formula for displacement during the \( n^{th} \) second:
\[
S_n = u + \frac{a}{2} \cdot (2n - 1)
\]
Where:
- \( S_n \) is the distance covered in the \( n^{th} \) second
- \( u \) is the initial velocity
- \( a \) is the acceleration (which is \( -g \) here)
For the second second (\( n = 2 \)):
\[
S_2 = 20 + \frac{-10}{2} \cdot (2 \cdot 2 - 1)
\]
Calculating:
\[
S_2 = 20 - 5 \cdot 3 = 20 - 15 = 5 \, \text{m}
\]
### Step 3: Check the Second Option (Displacement in Second and Third Seconds)
We need to find the displacement in the third second (\( n = 3 \)):
\[
S_3 = u + \frac{a}{2} \cdot (2n - 1)
\]
For \( n = 3 \):
\[
S_3 = 20 + \frac{-10}{2} \cdot (2 \cdot 3 - 1)
\]
Calculating:
\[
S_3 = 20 - 5 \cdot 5 = 20 - 25 = -5 \, \text{m}
\]
The displacement in the second second is \( 5 \, \text{m} \) (upward), and in the third second, it is \( -5 \, \text{m} \) (downward). Therefore, the displacements are not equal.
### Step 4: Check the Third Option (Distance Covered in Second and Third Seconds)
The distance covered is the magnitude of the displacement:
- Distance in the second second: \( |S_2| = 5 \, \text{m} \)
- Distance in the third second: \( |S_3| = 5 \, \text{m} \)
Thus, the distances are equal, confirming that the third option is correct.
### Step 5: Calculate the Average Speed
To find the average speed over the total time of 4 seconds, we first need to calculate the total distance covered in each second:
1. **First Second** (\( n = 1 \)):
\[
S_1 = 20 + \frac{-10}{2} \cdot (2 \cdot 1 - 1) = 20 - 5 = 15 \, \text{m}
\]
2. **Second Second** (\( n = 2 \)):
\[
S_2 = 5 \, \text{m} \quad \text{(from previous calculation)}
\]
3. **Third Second** (\( n = 3 \)):
\[
S_3 = 5 \, \text{m} \quad \text{(from previous calculation)}
\]
4. **Fourth Second** (\( n = 4 \)):
\[
S_4 = 20 + \frac{-10}{2} \cdot (2 \cdot 4 - 1) = 20 - 5 \cdot 7 = 20 - 35 = -15 \, \text{m}
\]
### Total Distance Calculation
Total distance \( S \):
\[
S = S_1 + S_2 + S_3 + |S_4| = 15 + 5 + 5 + 15 = 40 \, \text{m}
\]
### Average Speed Calculation
Average speed \( V_{avg} \):
\[
V_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{40 \, \text{m}}{4 \, \text{s}} = 10 \, \text{m/s}
\]
### Conclusion
- Distance covered in the second second: \( 5 \, \text{m} \) (Correct)
- Displacements in second and third seconds are not equal (Incorrect)
- Distances in second and third seconds are equal (Correct)
- Average speed: \( 10 \, \text{m/s} \)
