A ball is thrown vertcally upwards with a velocity `v` and an intial kinetic energy `E`. When half way to the top of its flight, it can have velocity and kinetic energy respectively of `:`

Options

A. `v/2,E/2`
B. `v/sqrt2,E/2`
C. `v/4,E/2`
D. `v/4,E/sqrt2`

Explanation

To solve the problem, we need to analyze the motion of the ball thrown vertically upwards. Let's break down the solution step by step.

### Step 1: Determine the maximum height (h) the ball reaches.
When a ball is thrown upwards with an initial velocity \( v \), it will rise to a maximum height \( h \) where its final velocity becomes 0. We can use the equation of motion for this:

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

Here, \( u = v \) (initial velocity), and at the maximum height, the final velocity \( v = 0 \). Rearranging gives:

\[ 0 = v^2 - 2gh \implies h = \frac{v^2}{2g} \]

### Step 2: Calculate the height at halfway point.
The halfway point to the maximum height is \( \frac{h}{2} \):

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

### Step 3: Use the kinematic equation to find the velocity at halfway point.
Now we will find the velocity of the ball when it has risen to \( \frac{h}{2} \). We can use the same kinematic equation:

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

Substituting \( u = v \) and \( \frac{h}{2} = \frac{v^2}{4g} \):

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

This simplifies to:

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

### Step 4: Calculate the velocity at halfway point.
Taking the square root gives:

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

### Step 5: Calculate the kinetic energy at halfway point.
The initial kinetic energy \( E \) when the ball was thrown is given by:

\[ E = \frac{1}{2} mv^2 \]

At the halfway point, the kinetic energy \( E' \) can be calculated using the velocity we just found:

\[ E' = \frac{1}{2} m v_n^2 = \frac{1}{2} m \left(\frac{v^2}{2}\right) = \frac{1}{4} mv^2 \]

Since \( E = \frac{1}{2} mv^2 \), we can express \( E' \) in terms of \( E \):

\[ E' = \frac{1}{4} mv^2 = \frac{1}{2} \cdot \frac{1}{2} mv^2 = \frac{E}{2} \]

### Conclusion:
Thus, at halfway to the top of its flight, the ball has:

- Velocity: \( \frac{v}{\sqrt{2}} \)
- Kinetic Energy: \( \frac{E}{2} \)

### Final Answer:
The velocity and kinetic energy at halfway to the top of its flight are \( \frac{v}{\sqrt{2}} \) and \( \frac{E}{2} \), respectively.

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