A plank fitted with a gun is moving on a horizontal surface with speed of `4m//s` along the positive x-axis. The z-axis is in vertically upward direction. The mass of the plank including the mass of the gun is 50 kg. When the plank reaches the origin, a shell of mass 10 kg is fired at an angle of `60^@` with the positive x-axis with a speed of `v = 20m//s` with respect to the gun in x-z plane. Find the position vector of the shell at t=2 s after firing it. Take `g = 9.8 m//s^2`.

Explanation

To solve the problem, we need to find the position vector of the shell at \( t = 2 \) seconds after it is fired. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the Initial Conditions
- The plank (including the gun) is moving with a speed of \( 4 \, \text{m/s} \) along the positive x-axis.
- The shell is fired at an angle of \( 60^\circ \) with respect to the positive x-axis with a speed of \( 20 \, \text{m/s} \) relative to the gun.

### Step 2: Determine the Velocity of the Shell
The velocity of the shell can be determined by combining the velocity of the plank and the velocity of the shell relative to the plank.

1. **Velocity of the shell in the x-direction**:
\[ v_{x} = v_{plank} + v_{shell} \cdot \cos(60^\circ) \]
\[ v_{x} = 4 \, \text{m/s} + 20 \cdot \cos(60^\circ) = 4 + 20 \cdot \frac{1}{2} = 4 + 10 = 14 \, \text{m/s} \]

2. **Velocity of the shell in the z-direction**:
\[ v_{z} = v_{shell} \cdot \sin(60^\circ) \]
\[ v_{z} = 20 \cdot \sin(60^\circ) = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]

### Step 3: Calculate the Position of the Shell at \( t = 2 \) seconds
Now, we can calculate the position of the shell in both x and z directions at \( t = 2 \) seconds.

1. **Position in the x-direction**:
\[ x(t) = v_{x} \cdot t = 14 \cdot 2 = 28 \, \text{m} \]

2. **Position in the z-direction**:
Using the second equation of motion for vertical motion:
\[ z(t) = v_{z} \cdot t - \frac{1}{2} g t^2 \]
\[ z(t) = (10\sqrt{3}) \cdot 2 - \frac{1}{2} \cdot 9.8 \cdot (2^2) \]
\[ z(t) = 20\sqrt{3} - \frac{1}{2} \cdot 9.8 \cdot 4 \]
\[ z(t) = 20\sqrt{3} - 19.6 \]
Approximating \( \sqrt{3} \approx 1.732 \):
\[ z(t) \approx 20 \cdot 1.732 - 19.6 \approx 34.64 - 19.6 \approx 15.04 \, \text{m} \]

### Step 4: Write the Position Vector
The position vector \( \vec{r} \) of the shell at \( t = 2 \) seconds is given by:
\[ \vec{r} = x(t) \hat{i} + z(t) \hat{k} = 28 \hat{i} + 15.04 \hat{k} \, \text{m} \]

### Final Answer
The position vector of the shell at \( t = 2 \) seconds after firing it is:
\[ \vec{r} = 28 \hat{i} + 15.04 \hat{k} \, \text{m} \]