A particle is projected upwards with initial velocity u. assume acceleration due to gravity is `10m//s^(2)`. It is found that particle covers 5m in last second before reaching the maximum height. What can be the possible value of u?

Options

A. 20m/s
B. 35m/s
C. 5m/s
D. 40m/s

Explanation

To solve the problem, we need to determine the initial velocity \( u \) of a particle projected upwards, given that it covers a distance of 5 m in the last second before reaching its maximum height. The acceleration due to gravity is \( g = 10 \, \text{m/s}^2 \).

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The particle is projected upwards with an initial velocity \( u \).
- It moves upwards until it reaches its maximum height, where its velocity becomes zero.
- The distance covered in the last second before reaching the maximum height is given as 5 m.

2. **Using the Formula for Distance in the nth Second**:
- The distance covered in the nth second can be calculated using the formula:
\[ S_n = u + \frac{a}{2} \cdot (2n - 1) \]
- Here, \( S_n \) is the distance covered in the nth second, \( u \) is the initial velocity, \( a \) is the acceleration (which is \(-g\) or \(-10 \, \text{m/s}^2\) in this case), and \( n \) is the total time taken to reach the maximum height.

3. **Setting Up the Equation**:
- Since the distance covered in the last second (which is the nth second) is 5 m, we can write:
\[ 5 = u + \frac{-10}{2} \cdot (2n - 1) \]
- Simplifying this gives:
\[ 5 = u - 5(2n - 1) \]
- Rearranging gives:
\[ 5 = u - 10n + 5 \]
- Thus:
\[ u = 10n \]

4. **Finding the Time to Reach Maximum Height**:
- The time \( n \) taken to reach the maximum height can also be found using the equation:
\[ v = u + at \]
- At maximum height, the final velocity \( v = 0 \):
\[ 0 = u - 10n \]
- This implies:
\[ u = 10n \]

5. **Substituting for \( n \)**:
- From the previous step, we have \( u = 10n \). We can substitute this into the distance equation:
\[ 5 = 10n - 5(2n - 1) \]
- Simplifying gives:
\[ 5 = 10n - 10n + 5 \]
- This equation is satisfied for any \( n \geq 1 \).

6. **Finding Possible Values of \( u \)**:
- Since \( n \) must be greater than or equal to 1, we can substitute values for \( n \):
- If \( n = 1 \), then \( u = 10 \times 1 = 10 \, \text{m/s} \).
- If \( n = 2 \), then \( u = 10 \times 2 = 20 \, \text{m/s} \).
- If \( n = 3 \), then \( u = 10 \times 3 = 30 \, \text{m/s} \).
- If \( n = 4 \), then \( u = 10 \times 4 = 40 \, \text{m/s} \).

### Conclusion:
The possible values of \( u \) are \( 10 \, \text{m/s}, 20 \, \text{m/s}, 30 \, \text{m/s}, \) and \( 40 \, \text{m/s} \).