Two shells are fired from cannon with speed `u` each, at angles of `alpha` of `beta` respectively with the horizontal.The time interval between the shots is `T`.They collide in mid air after time `t` from the first shot.Which of the following conditions must be satisfied?

Options

A. `alphagtbeta`
B. `t cos alpha=(t-T) cos beta`
C. `(t-T) cos alpha =tcos beta`
D. `(u sin alpha)t-(1//2) g(t)^(2) = (u sin beta)(t-T)-(1//2)g(t-T)^(2)`

Answer

To solve the problem, we need to analyze the motion of the two shells fired from the cannon. Let's break down the steps to derive the conditions that must be satisfied for the shells to collide in mid-air.

### Step 1: Determine the horizontal and vertical components of the velocities
For both shells, the horizontal and vertical components of their velocities can be expressed as follows:

1. **First shell (fired at angle α)**:
- Horizontal component: \( u_x = u \cos \alpha \)
- Vertical component: \( v_y = u \sin \alpha \)

2. **Second shell (fired at angle β)**:
- Horizontal component: \( u_x = u \cos \beta \)
- Vertical component: \( v_y = u \sin \beta \)

### Step 2: Calculate the horizontal distance traveled by both shells
After time \( t \) from the first shot, the horizontal distance traveled by the first shell is:
\[ x_1 = u \cos \alpha \cdot t \]

The second shell is fired after a time interval \( T \), so by the time the first shell has been in the air for \( t \), the second shell has been in the air for \( t - T \). The horizontal distance traveled by the second shell is:
\[ x_2 = u \cos \beta \cdot (t - T) \]

### Step 3: Set the horizontal distances equal for collision
For the shells to collide, they must be at the same horizontal position:
\[ u \cos \alpha \cdot t = u \cos \beta \cdot (t - T) \]
Dividing both sides by \( u \) (assuming \( u \neq 0 \)):
\[ \cos \alpha \cdot t = \cos \beta \cdot (t - T) \]

### Step 4: Calculate the vertical distance traveled by both shells
The vertical distance for the first shell after time \( t \) is:
\[ y_1 = u \sin \alpha \cdot t - \frac{1}{2} g t^2 \]

For the second shell, after time \( t - T \):
\[ y_2 = u \sin \beta \cdot (t - T) - \frac{1}{2} g (t - T)^2 \]

### Step 5: Set the vertical distances equal for collision
For the shells to collide, their vertical positions must also be equal:
\[ u \sin \alpha \cdot t - \frac{1}{2} g t^2 = u \sin \beta \cdot (t - T) - \frac{1}{2} g (t - T)^2 \]

### Step 6: Analyze the conditions
From the horizontal distance equation, we have:
\[ \cos \alpha \cdot t = \cos \beta \cdot (t - T) \]

From the vertical distance equation, we can derive another condition.

### Conclusion
The conditions that must be satisfied for the shells to collide are:
1. \( \cos \alpha \cdot t = \cos \beta \cdot (t - T) \)
2. The vertical distance condition derived from the second equation.