A cannon fires successively two shells from the same point with velocity `V_(0)=250m//s`, the first at the angle `theta_(1)=60^(@)` and the second at the angle `theta_(2)=45^(@)` to the horizontal, the azimuth being the same. Neglecting the air drag, find the approximate time interval between firings leading to the collision of the shells `(g=9.8m//s^(2))`

Options

A. `11 sec`
B. `6 sec`
C. `15 sec`
D. `5 sec`

Explanation

To solve the problem of finding the time interval between the firings of two shells from a cannon, we will follow these steps:

### Step 1: Determine the horizontal components of the velocities
The horizontal component of the velocity for each shell can be calculated using the formula:
- For the first shell (fired at an angle of 60 degrees):
\[ V_{0x1} = V_0 \cos(60^\circ) = 250 \cdot \frac{1}{2} = 125 \, \text{m/s} \]
- For the second shell (fired at an angle of 45 degrees):
\[ V_{0x2} = V_0 \cos(45^\circ) = 250 \cdot \frac{1}{\sqrt{2}} \approx 176.78 \, \text{m/s} \]

### Step 2: Set up the equations for horizontal motion
Since both shells must collide at the same horizontal distance, we can set up the equations:
\[ x_1 = V_{0x1} t_1 = 125 t_1 \]
\[ x_2 = V_{0x2} t_2 = 176.78 t_2 \]
Setting \(x_1 = x_2\):
\[ 125 t_1 = 176.78 t_2 \]
From this, we can express \(t_1\) in terms of \(t_2\):
\[ t_1 = \frac{176.78}{125} t_2 \approx 1.414 t_2 \]

### Step 3: Determine the vertical components of the velocities
Next, we calculate the vertical components of the velocities:
- For the first shell:
\[ V_{0y1} = V_0 \sin(60^\circ) = 250 \cdot \frac{\sqrt{3}}{2} \approx 216.51 \, \text{m/s} \]
- For the second shell:
\[ V_{0y2} = V_0 \sin(45^\circ) = 250 \cdot \frac{1}{\sqrt{2}} \approx 176.78 \, \text{m/s} \]

### Step 4: Set up the equations for vertical motion
Using the vertical motion equation \(y = V_{0y} t - \frac{1}{2} g t^2\):
- For the first shell:
\[ y_1 = V_{0y1} t_1 - \frac{1}{2} g t_1^2 \]
- For the second shell:
\[ y_2 = V_{0y2} t_2 - \frac{1}{2} g t_2^2 \]
Setting \(y_1 = y_2\):
\[ 216.51 t_1 - \frac{1}{2} g t_1^2 = 176.78 t_2 - \frac{1}{2} g t_2^2 \]

### Step 5: Substitute \(t_1\) in terms of \(t_2\)
Substituting \(t_1 = 1.414 t_2\) into the vertical motion equation:
\[ 216.51 (1.414 t_2) - \frac{1}{2} g (1.414 t_2)^2 = 176.78 t_2 - \frac{1}{2} g t_2^2 \]
This simplifies to:
\[ 306.67 t_2 - \frac{1}{2} g (2 t_2^2) = 176.78 t_2 - \frac{1}{2} g t_2^2 \]

### Step 6: Rearranging and solving for \(t_2\)
Rearranging gives:
\[ 306.67 t_2 - 176.78 t_2 = \frac{1}{2} g t_2^2 \]
\[ 129.89 t_2 = \frac{1}{2} g t_2^2 \]
Dividing both sides by \(t_2\) (assuming \(t_2 \neq 0\)):
\[ 129.89 = \frac{1}{2} g t_2 \]
Substituting \(g = 9.8 \, \text{m/s}^2\):
\[ 129.89 = \frac{1}{2} \cdot 9.8 \cdot t_2 \]
\[ t_2 = \frac{129.89 \cdot 2}{9.8} \approx 26.5 \, \text{s} \]

### Step 7: Finding \(t_1\)
Now substituting back to find \(t_1\):
\[ t_1 = 1.414 t_2 \approx 1.414 \cdot 26.5 \approx 37.4 \, \text{s} \]

### Step 8: Calculate the time interval
The time interval between the firings is:
\[ \Delta t = t_1 - t_2 \approx 37.4 - 26.5 \approx 10.9 \, \text{s} \]

### Final Answer
The approximate time interval between the firings leading to the collision of the shells is **11 seconds**.