Maximum height attained from the point of projection is:

Options

A. 1.25 m
B. 12.5 m
C. 2025 m
D. None of these

Explanation

To find the maximum height attained from the point of projection for a projectile, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Time of flight (T) = 1 second
- Acceleration due to gravity (g) = 10 m/s² (standard value)
- We need to find the initial vertical component of the velocity (u_y) and the initial horizontal component of the velocity (u_x).

2. **Use the Time of Flight Formula:**
The time of flight for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
Rearranging this formula to find \( u \sin \theta \):
\[ u \sin \theta = \frac{gT}{2} \]
Substituting the known values:
\[ u \sin \theta = \frac{10 \times 1}{2} = 5 \text{ m/s} \]

3. **Calculate the Range:**
The range (R) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Given that the range is 4 meters, we can rearrange this to find \( u \):
\[ 4 = \frac{u^2 \sin 2\theta}{10} \]
This implies:
\[ u^2 \sin 2\theta = 40 \]

4. **Find the Horizontal Component of Velocity:**
The horizontal component of the initial velocity is related to the range:
\[ R = u_x \cdot T \]
Thus:
\[ u_x = \frac{R}{T} = \frac{4}{1} = 4 \text{ m/s} \]

5. **Find the Vertical Component of Velocity:**
We already found \( u \sin \theta = 5 \text{ m/s} \). Now we need to find \( u \):
\[ u = \sqrt{u_x^2 + u_y^2} \]
We have \( u_x = 4 \text{ m/s} \) and \( u_y = 5 \text{ m/s} \):
\[ u = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \text{ m/s} \]

6. **Calculate Maximum Height:**
The maximum height (h) attained by the projectile can be calculated using the formula:
\[ h = \frac{u_y^2}{2g} \]
Substituting \( u_y = 5 \text{ m/s} \) and \( g = 10 \text{ m/s}^2 \):
\[ h = \frac{5^2}{2 \times 10} = \frac{25}{20} = 1.25 \text{ m} \]

### Final Answer:
The maximum height attained from the point of projection is **1.25 meters**.