A body of mass m thrown vertically upwards attains a maximum height h. At height will its kinetic energy be 75% of its intial value?

Options

A. `(h)/(6)`
B. `(h)/(5)`
C. `(h)/(4)`
D. `(h)/(3)`

Explanation

To solve the problem step by step, we need to find the height at which the kinetic energy of a body thrown vertically upwards is 75% of its initial kinetic energy. Let's break down the solution:

### Step 1: Understand the Initial Conditions
The body of mass \( m \) is thrown upwards with an initial velocity \( u \) and reaches a maximum height \( h \). At this maximum height, all the kinetic energy is converted into potential energy.

### Step 2: Write the Initial Kinetic Energy
The initial kinetic energy \( KE_i \) when the body is thrown is given by:
\[ KE_i = \frac{1}{2} m u^2 \]

### Step 3: Write the Potential Energy at Maximum Height
At the maximum height \( h \), the kinetic energy becomes zero, and all of it is converted into potential energy \( PE \):
\[ PE = mgh \]
By the conservation of energy, we have:
\[ \frac{1}{2} m u^2 = mgh \]

### Step 4: Find the Velocity at 75% of Initial Kinetic Energy
We need to find the height at which the kinetic energy is 75% of the initial kinetic energy:
\[ KE = 0.75 \times KE_i = 0.75 \times \frac{1}{2} m u^2 = \frac{3}{8} m u^2 \]

### Step 5: Relate Kinetic Energy to Velocity
The kinetic energy at a certain height \( x \) can also be expressed as:
\[ KE = \frac{1}{2} m v^2 \]
Setting this equal to the 75% kinetic energy we found:
\[ \frac{1}{2} m v^2 = \frac{3}{8} m u^2 \]
Cancelling \( m \) from both sides gives:
\[ \frac{1}{2} v^2 = \frac{3}{8} u^2 \]
Multiplying both sides by 2:
\[ v^2 = \frac{3}{4} u^2 \]
Taking the square root:
\[ v = \frac{\sqrt{3}}{2} u \]

### Step 6: Use Energy Conservation to Find Height \( x \)
At height \( x \), the total mechanical energy is conserved. The total energy at height \( h \) is:
\[ mgh = KE + PE \]
At height \( x \):
\[ mgh = \frac{1}{2} m v^2 + mgx \]
Substituting \( v^2 = \frac{3}{4} u^2 \):
\[ mgh = \frac{1}{2} m \left(\frac{3}{4} u^2\right) + mgx \]
Substituting \( u^2 = 2gh \) from the initial energy conservation:
\[ mgh = \frac{1}{2} m \left(\frac{3}{4} \cdot 2gh\right) + mgx \]
This simplifies to:
\[ mgh = \frac{3}{4} mgh + mgx \]
Cancelling \( mg \) from both sides:
\[ h = \frac{3}{4} h + x \]
Rearranging gives:
\[ x = h - \frac{3}{4} h = \frac{1}{4} h \]

### Conclusion
Thus, the height at which the kinetic energy is 75% of its initial value is:
\[ x = \frac{h}{4} \]

### Final Answer
The correct option is \( \frac{h}{4} \).

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