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 upward with an initial speed \( u \) and determine the time \( T \) it takes to reach the maximum height \( H \).

### Step-by-Step Solution:

1. **Understanding the Motion**:
When a body is projected upwards, it moves against the force of gravity. The acceleration due to gravity \( g \) acts downwards, which means it will decelerate the body until it reaches its maximum height.

2. **Using the Equations of Motion**:
At the maximum height, the final velocity \( v \) of the body becomes zero. We can use the first equation of motion:
\[ v = u - gT \]
At maximum height, \( v = 0 \), so we can set the equation to:
\[ 0 = u - gT \]

3. **Rearranging the Equation**:
Rearranging the above equation gives us:
\[ gT = u \]
Therefore, we can express the time \( T \) taken to reach the maximum height as:
\[ T = \frac{u}{g} \]

4. **Finding the Maximum Height \( H \)**:
We can also find the maximum height \( H \) using the second equation of motion:
\[ H = uT - \frac{1}{2}gT^2 \]
Substituting \( T = \frac{u}{g} \) into the equation:
\[ H = u\left(\frac{u}{g}\right) - \frac{1}{2}g\left(\frac{u}{g}\right)^2 \]
Simplifying this gives:
\[ H = \frac{u^2}{g} - \frac{1}{2}g\left(\frac{u^2}{g^2}\right) \]
\[ H = \frac{u^2}{g} - \frac{u^2}{2g} = \frac{u^2}{2g} \]

5. **Conclusion**:
The time taken \( T \) to reach the maximum height \( H \) is given by:
\[ T = \frac{u}{g} \]
and the maximum height \( H \) is given by:
\[ H = \frac{u^2}{2g} \]