A body thrown vertically up with velocity u reaches the maximum height h after T seconds. Which of the following statements is true ?

Options

A. At a height `h/2` from the ground its velocity is `u/2`
B. At a time T its velocity is u
C. At a time 2T its velocity is u directed downwards
D. none of the above

Answer

To solve the problem, we need to analyze the motion of a body thrown vertically upwards with an initial velocity \( u \) until it reaches its maximum height \( h \) after \( T \) seconds. We will evaluate the given statements one by one.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The body is thrown upwards with an initial velocity \( u \).
- It reaches a maximum height \( h \) after \( T \) seconds.
- At maximum height, the final velocity \( v = 0 \).

2. **Using the Kinematic Equation**:
- We can use the kinematic equation:
\[ v^2 = u^2 + 2a s \]
- Here, \( v = 0 \) (at maximum height), \( u = u \), \( a = -g \) (acceleration due to gravity), and \( s = h \).
- Substituting these values, we get:
\[ 0 = u^2 - 2gh \]
- Rearranging gives:
\[ 2gh = u^2 \quad \text{(1)} \]

3. **Finding Velocity at Height \( \frac{h}{2} \)**:
- Now, we want to find the velocity at a height \( \frac{h}{2} \).
- Using the same kinematic equation:
\[ v^2 = u^2 + 2a s \]
- Here, \( s = \frac{h}{2} \), so:
\[ v^2 = u^2 - 2g\left(\frac{h}{2}\right) \]
- Substituting \( 2gh = u^2 \) from equation (1):
\[ v^2 = u^2 - gh \]
- Since \( gh = \frac{u^2}{2} \) (from equation (1)), we substitute:
\[ v^2 = u^2 - \frac{u^2}{2} = \frac{u^2}{2} \]
- Therefore, taking the square root:
\[ v = \frac{u}{\sqrt{2}} \]

4. **Evaluating the Statements**:
- **Statement 1**: "At a height \( \frac{h}{2} \), its velocity is \( \frac{u}{2} \)".
- This is **false** because we found \( v = \frac{u}{\sqrt{2}} \).
- **Statement 2**: "At time \( T \), its velocity is \( u \)".
- This is **false** because at time \( T \), the body reaches maximum height and \( v = 0 \).
- **Statement 3**: "At time \( 2T \), its velocity is \( u \) directed downwards".
- After reaching maximum height, the body falls back down. At time \( 2T \), it will have the same speed \( u \) but in the downward direction, making this statement **true**.
- **Statement 4**: "None of the above".
- This is **false** since statement 3 is true.

### Conclusion:
The correct statement is that at time \( 2T \), its velocity is \( u \) directed downwards.