A particle is projected vertically upwards in vacuum with a speed `u`.

Options

A. When it rises to half its maximum height, its speed becomes `u//2`.
B. When it rises to half its maximum height, its speed becomes `u//sqrt(2)`.
C. The time taken to rise to half its maximum height is half the time taken to reach its maximum height.
D. The time taken to rise to three`-` fourth of its maximum height is half the time taken to reach its maximum height.

Explanation

To solve the problem of a particle projected vertically upwards in a vacuum with an initial speed \( u \), we will analyze the motion using the equations of motion under uniform acceleration due to gravity.

### Step-by-step Solution:

1. **Determine the maximum height (h)**:
The maximum height reached by the particle can be calculated using the equation:
\[ v^2 = u^2 - 2gh \]
At maximum height, the final velocity \( v = 0 \). Therefore, we can rearrange the equation:
\[ 0 = u^2 - 2gh \implies h = \frac{u^2}{2g} \]

2. **Calculate the speed at half the maximum height**:
The height at half the maximum height is:
\[ h/2 = \frac{u^2}{4g} \]
Now, we can use the same equation of motion to find the speed \( v \) at this height:
\[ v^2 = u^2 - 2g\left(\frac{u^2}{4g}\right) \]
Simplifying this:
\[ v^2 = u^2 - \frac{u^2}{2} = \frac{u^2}{2} \]
Therefore, the speed at half the maximum height is:
\[ v = \frac{u}{\sqrt{2}} \]

3. **Calculate the time taken to reach maximum height (T)**:
The time taken to reach the maximum height can be calculated using:
\[ v = u - gt \]
Setting \( v = 0 \) at maximum height:
\[ 0 = u - gT \implies T = \frac{u}{g} \]

4. **Calculate the time taken to reach half the maximum height**:
Using the speed at half the maximum height \( v = \frac{u}{\sqrt{2}} \):
\[ \frac{u}{\sqrt{2}} = u - g t \]
Rearranging gives:
\[ g t = u - \frac{u}{\sqrt{2}} = u\left(1 - \frac{1}{\sqrt{2}}\right) \]
Thus,
\[ t = \frac{u\left(1 - \frac{1}{\sqrt{2}}\right)}{g} \]
This time is not half of \( T \).

5. **Calculate the time taken to rise to three-fourths of maximum height**:
The height at three-fourths of maximum height is:
\[ \frac{3h}{4} = \frac{3u^2}{8g} \]
Using the equation of motion:
\[ v^2 = u^2 - 2g\left(\frac{3u^2}{8g}\right) \]
Simplifying gives:
\[ v^2 = u^2 - \frac{3u^2}{4} = \frac{u^2}{4} \implies v = \frac{u}{2} \]
Now, using the time equation:
\[ \frac{u}{2} = u - gt \implies gt = u - \frac{u}{2} = \frac{u}{2} \]
Thus,
\[ t = \frac{u}{2g} \]
This time \( t \) is indeed half of \( T \).

### Conclusion:
- The correct options are:
- The speed at half the maximum height is \( \frac{u}{\sqrt{2}} \).
- The time taken to rise to three-fourths of its maximum height is half the time taken to reach maximum height.