A body is projected vertically up with u. Its velocity at half its maximum height is

Options

A. `u/2`
B. `u^2/2`
C. `sqrt(2u)`
D. `u/sqrt2`

Answer

To find the velocity of a body projected vertically upwards at half its maximum height, we can follow these steps:

### Step 1: Determine the maximum height (h)
When a body is projected upwards with an initial velocity \( u \), it will reach a maximum height \( h \) where its final velocity becomes zero. We can use the third equation of motion:

\[ v^2 = u^2 + 2a s \]

Here, \( v = 0 \) (at maximum height), \( u \) is the initial velocity, \( a = -g \) (acceleration due to gravity, acting downwards), and \( s = h \) (the maximum height).

Substituting these values, we get:

\[ 0 = u^2 - 2gh \]

Rearranging gives:

\[ h = \frac{u^2}{2g} \]

### Step 2: Find the height at half maximum height
Now, we need to find the velocity at half of the maximum height, which is \( \frac{h}{2} \):

\[ \frac{h}{2} = \frac{1}{2} \cdot \frac{u^2}{2g} = \frac{u^2}{4g} \]

### Step 3: Use the third equation of motion again
Now, we will use the third equation of motion again to find the velocity \( v \) at this height \( \frac{h}{2} \):

\[ v^2 = u^2 + 2a s \]

Here, \( s = \frac{u^2}{4g} \) and \( a = -g \). Substituting these values, we have:

\[ v^2 = u^2 - 2g \left(\frac{u^2}{4g}\right) \]

This simplifies to:

\[ v^2 = u^2 - \frac{u^2}{2} = \frac{u^2}{2} \]

### Step 4: Calculate the velocity
Taking the square root to find \( v \):

\[ v = \sqrt{\frac{u^2}{2}} = \frac{u}{\sqrt{2}} \]

### Conclusion
Thus, the velocity of the body at half its maximum height is:

\[ v = \frac{u}{\sqrt{2}} \]