A particle is projected vertically upwards with a speed of `16ms^-1`. After some time, when it again passes through the point of projection, its speed is found to be `8ms^-1`. It is known that the work done by air resistance is same during upward and downward motion. Then the maximum height attained by the particle is (take `g=10ms^-2`)

Options

A. (a) `8m`
B. (b) `4.8m`
C. (c) `17.6m`
D. (d) `12.8m`

Explanation

To solve the problem step by step, we will use the work-energy principle and the information given about the motion of the particle.

### Step 1: Understand the Problem
A particle is projected vertically upwards with an initial speed \( u = 16 \, \text{m/s} \). When it returns to the point of projection, it has a speed of \( v = 8 \, \text{m/s} \). We need to find the maximum height \( H \) attained by the particle, considering that the work done by air resistance is the same during both the upward and downward motion.

### Step 2: Apply the Work-Energy Principle
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. We can set up two equations for the upward and downward motion.

#### Upward Motion (from point A to point B):
- Initial kinetic energy at point A:
\[ KE_i = \frac{1}{2} m u^2 = \frac{1}{2} m (16^2) = \frac{1}{2} m (256) = 128m \]
- Final kinetic energy at point B (at maximum height):
\[ KE_f = 0 \, \text{(at maximum height)} \]
- Work done against gravity:
\[ W_g = -mgh \]
- Work done by air resistance:
\[ W_d = -W_d \, \text{(opposite to the direction of motion)} \]

Using the work-energy theorem:
\[ W_d + W_g = KE_f - KE_i \]
Substituting the values:
\[ -W_d - mgh = 0 - 128m \]
This simplifies to:
\[ -W_d = 128m + mgh \quad \text{(Equation 1)} \]

#### Downward Motion (from point B to point C):
- Initial kinetic energy at point B:
\[ KE_i = 0 \, \text{(at maximum height)} \]
- Final kinetic energy at point C:
\[ KE_f = \frac{1}{2} m v^2 = \frac{1}{2} m (8^2) = \frac{1}{2} m (64) = 32m \]
- Work done against gravity:
\[ W_g = mgh \, \text{(positive work done by gravity)} \]
- Work done by air resistance:
\[ W_d = -W_d \, \text{(opposite to the direction of motion)} \]

Using the work-energy theorem:
\[ W_d + W_g = KE_f - KE_i \]
Substituting the values:
\[ -W_d + mgh = 32m - 0 \]
This simplifies to:
\[ mgh - W_d = 32m \quad \text{(Equation 2)} \]

### Step 3: Combine the Equations
Now we have two equations:
1. \( -W_d = 128m + mgh \)
2. \( mgh - W_d = 32m \)

From Equation 1, we can express \( W_d \):
\[ W_d = -128m - mgh \]

Substituting this into Equation 2:
\[ mgh - (-128m - mgh) = 32m \]
This simplifies to:
\[ mgh + 128m + mgh = 32m \]
\[ 2mgh + 128m = 32m \]
\[ 2mgh = 32m - 128m \]
\[ 2mgh = -96m \]
Dividing both sides by \( 2m \):
\[ gh = -48 \]
Since \( g = 10 \, \text{m/s}^2 \):
\[ h = \frac{-48}{10} = -4.8 \, \text{m} \]
This indicates a mistake in the signs or assumptions. Let's revisit the equations.

### Step 4: Correcting the Equations
We need to consider the absolute values and the direction of work done. The correct approach would be to set the work done by air resistance equal during both motions.

Using the conservation of energy:
\[ \text{Initial Energy} = \text{Final Energy} + \text{Work done by air resistance} \]

### Final Calculation for Maximum Height
Using the final speeds and the work done:
\[ H = \frac{u^2 - v^2}{2g} \]
Substituting the values:
\[ H = \frac{16^2 - 8^2}{2 \times 10} \]
\[ H = \frac{256 - 64}{20} = \frac{192}{20} = 9.6 \, \text{m} \]

### Conclusion
The maximum height attained by the particle is \( 9.6 \, \text{m} \).