A body is projected up with a speed `u` and the time taken by it is T to reach the maximum height H. Pich out the correct statement

Options

A. it reaches `(H)/(2)` in `(T)/(2)` sec
B. It acquires velocity `(u)/(2)` in (T)/(2)` sec
C. Its velocity is `(u)/(2)` at `(H)/(2)`
D. same velocity at 2T

Explanation

To solve the problem, we need to analyze the motion of a body projected upwards with an initial speed \( u \) under the influence of gravity. The time taken to reach the maximum height is given as \( T \).

### Step-by-Step Solution:

1. **Understanding the Motion**:
- When a body is projected upwards, it moves against the gravitational force until it reaches its maximum height \( H \). At this point, its velocity becomes zero.

2. **Using the Equations of Motion**:
- We can use the following equations of motion:
1. \( s = ut + \frac{1}{2} a t^2 \)
2. \( v - u = at \)
3. \( v^2 - u^2 = 2as \)

Here, \( s \) is the displacement, \( u \) is the initial velocity, \( v \) is the final velocity, \( a \) is the acceleration (which is \( -g \) for upward motion), and \( t \) is the time.

3. **Finding the Maximum Height \( H \)**:
- At maximum height, the final velocity \( v = 0 \).
- Using the second equation:
\[ 0 - u = -gT \implies u = gT \]
- Now, substituting \( u \) into the first equation to find \( H \):
\[ H = uT - \frac{1}{2}gT^2 = gT \cdot T - \frac{1}{2}gT^2 = gT^2 - \frac{1}{2}gT^2 = \frac{1}{2}gT^2 \]

4. **Analyzing the Options**:
- **Option 1**: "It reaches \( H/2 \) in time \( T/2 \)".
- Using the first equation for \( T/2 \):
\[ s = u \cdot \frac{T}{2} - \frac{1}{2}g\left(\frac{T}{2}\right)^2 = gT \cdot \frac{T}{2} - \frac{1}{2}g\frac{T^2}{4} = \frac{gT^2}{2} - \frac{gT^2}{8} = \frac{4gT^2 - gT^2}{8} = \frac{3gT^2}{8} \]
- Since \( H/2 = \frac{1}{4}gT^2 \), this option is incorrect.

- **Option 2**: "It acquires velocity \( u/2 \) in \( T/2 \) seconds".
- Using the second equation:
\[ v = u - g\frac{T}{2} = gT - g\frac{T}{2} = g\frac{T}{2} = \frac{u}{2} \]
- This option is correct.

- **Option 3**: "Its velocity is \( u/2 \) at \( H/2 \)".
- Using the third equation:
\[ v^2 = u^2 - 2g\left(\frac{H}{2}\right) = (gT)^2 - 2g\left(\frac{1}{2}gT^2\right) = g^2T^2 - g^2T^2 = 0 \]
- This option is incorrect.

- **Option 4**: "The same velocity at \( 2T \)".
- The velocity will be equal in magnitude but opposite in direction when it returns to the starting point. Thus, this option is incorrect.

### Conclusion:
The correct statement is **Option 2**: "It acquires velocity \( u/2 \) in \( T/2 \) seconds".