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, we can follow these steps:

### Step 1: Understand the relationship between time of flight, initial velocity, and angle of projection
The time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{u \sin \theta}{g} \]
where:
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 2: Rearrange the formula to find \( u \sin \theta \)
Given that the time of flight \( T \) is \( 1 \) second, we can rearrange the formula:
\[ u \sin \theta = g \cdot T \]
Substituting the known values:
\[ u \sin \theta = 10 \cdot 1 = 10 \, \text{m/s} \]

### Step 3: Use the formula for maximum height
The maximum height \( H \) attained by the projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

### Step 4: Substitute \( u \sin \theta \) into the height formula
From the previous step, we have \( u \sin \theta = 10 \, \text{m/s} \). To find \( u \sin^2 \theta \), we can use the identity:
\[ u \sin \theta = 10 \implies (u \sin \theta)^2 = 100 \]
Thus, we can express \( H \) as:
\[ H = \frac{(u \sin \theta)^2}{2g} = \frac{100}{2 \cdot 10} = \frac{100}{20} = 5 \, \text{m} \]

### Step 5: Calculate the maximum height
Now, substituting the values into the maximum height formula:
\[ H = \frac{10^2}{2 \cdot 10} = \frac{100}{20} = 5 \, \text{m} \]

### Step 6: Final answer
The maximum height attained from the point of projection is:
\[ \text{Maximum Height} = 5 \, \text{m} \]