A cannon of mass `1000 kg` located at the base of an inclined plane fires a shell of mass `50 kg` in horizontal direction with velocity `180 km//h`. The angle of inclination of the inclined plane with the horizontal is `45^(@)`. The coefficient of friction between the cannon and inclined plane is `0.5`. The maximum height, in metre, to which the cannon can ascend the inclined plane as a result of recoil is

Options

A. `5/6`
B. `5/24`
C. `5/12`
D. none of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the shell's velocity from km/h to m/s
The shell's velocity is given as \(180 \, \text{km/h}\). We need to convert this to meters per second (m/s) using the conversion factor \(1 \, \text{km/h} = \frac{1}{3.6} \, \text{m/s}\).

\[ v_{shell} = 180 \, \text{km/h} \times \frac{1 \, \text{m/s}}{3.6 \, \text{km/h}} = 50 \, \text{m/s} \]

### Step 2: Apply conservation of momentum
Before the cannon fires, its momentum is zero. After firing, the momentum of the shell and the cannon must be equal and opposite.

Let \(v\) be the recoil velocity of the cannon. The mass of the cannon is \(1000 \, \text{kg}\) and the mass of the shell is \(50 \, \text{kg}\).

Using conservation of momentum:

\[ 0 = (50 \, \text{kg} \times 50 \, \text{m/s}) + (1000 \, \text{kg} \times v) \]

Solving for \(v\):

\[ 1000v = -2500 \implies v = -\frac{2500}{1000} = -2.5 \, \text{m/s} \]

### Step 3: Calculate the effective velocity of the cannon along the incline
Since the cannon moves up the incline at an angle of \(45^\circ\), the effective velocity \(v_{cannon}\) along the incline is:

\[ v_{cannon} = v \cos(45^\circ) = -2.5 \times \frac{1}{\sqrt{2}} = -\frac{2.5}{\sqrt{2}} \approx -1.77 \, \text{m/s} \]

### Step 4: Calculate the frictional force
The frictional force \(F_f\) acting on the cannon is given by:

\[ F_f = \mu m g \cos(45^\circ) \]

Where:
- \(\mu = 0.5\)
- \(m = 1000 \, \text{kg}\)
- \(g = 10 \, \text{m/s}^2\)

Calculating \(F_f\):

\[ F_f = 0.5 \times 1000 \times 10 \times \frac{1}{\sqrt{2}} = \frac{5000}{\sqrt{2}} \approx 3535.53 \, \text{N} \]

### Step 5: Apply the work-energy principle
The work done against friction as the cannon moves up the incline is equal to the change in kinetic energy.

The initial kinetic energy of the cannon is:

\[ KE_{initial} = \frac{1}{2} m v_{cannon}^2 = \frac{1}{2} \times 1000 \times \left(-1.77\right)^2 \approx 1565.45 \, \text{J} \]

The work done against friction while moving up height \(h\) is:

\[ W = F_f \cdot h = \frac{5000}{\sqrt{2}} \cdot h \]

Setting the work done equal to the initial kinetic energy gives:

\[ \frac{5000}{\sqrt{2}} h = 1565.45 \]

### Step 6: Solve for height \(h\)
Rearranging the equation to find \(h\):

\[ h = \frac{1565.45 \cdot \sqrt{2}}{5000} \approx \frac{1565.45 \cdot 1.414}{5000} \approx \frac{2212.12}{5000} \approx 0.4424 \, \text{m} \]

### Final Answer
The maximum height to which the cannon can ascend the inclined plane as a result of recoil is approximately:

\[ \boxed{0.442 \, \text{m}} \]