A vertically projected body attains the maximum height in `6s`. The ratio of kinetic energy at the end of `3^(rd)` second to decrease in kinetic energy in the next three seconds is

Options

A. `1 : 1`
B. `1 : 3`
C. `3 : 1`
D. `9 : 1`

Explanation

To solve the problem step by step, we will follow the concepts of kinematics and kinetic energy.

### Step 1: Understanding the Motion
A body is projected vertically upwards and reaches its maximum height in 6 seconds. At maximum height, the final velocity (V) is 0 m/s.

### Step 2: Finding Initial Velocity (U)
Using the equation of motion:
\[ V = U - gt \]
At maximum height (t = 6 s):
\[ 0 = U - g \cdot 6 \]
Rearranging gives:
\[ U = 6g \]

### Step 3: Finding Velocity at the End of 3 Seconds (V3)
Now, we need to find the velocity at the end of 3 seconds:
Using the same equation of motion:
\[ V3 = U - g \cdot 3 \]
Substituting U:
\[ V3 = 6g - g \cdot 3 \]
\[ V3 = 6g - 3g = 3g \]

### Step 4: Calculating Kinetic Energy at the End of 3 Seconds
The kinetic energy (KE) at the end of 3 seconds is given by:
\[ KE = \frac{1}{2} m V^2 \]
Substituting V3:
\[ KE = \frac{1}{2} m (3g)^2 \]
\[ KE = \frac{1}{2} m \cdot 9g^2 = \frac{9mg^2}{2} \]

### Step 5: Finding Kinetic Energy at Maximum Height
At maximum height (after 6 seconds), the kinetic energy is:
\[ KE_{max height} = 0 \]
Thus, the decrease in kinetic energy from 3 seconds to maximum height is:
\[ \Delta KE = KE_{3s} - KE_{max height} \]
\[ \Delta KE = \frac{9mg^2}{2} - 0 = \frac{9mg^2}{2} \]

### Step 6: Finding the Ratio
Now we need to find the ratio of kinetic energy at the end of the 3rd second to the decrease in kinetic energy in the next three seconds:
\[ \text{Ratio} = \frac{KE_{3s}}{\Delta KE} = \frac{\frac{9mg^2}{2}}{\frac{9mg^2}{2}} = 1 \]

### Final Answer
The ratio of kinetic energy at the end of the 3rd second to the decrease in kinetic energy in the next three seconds is **1:1**.

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