A shot is fired at an angle `alpha` to the horizontal up a hill of inclination `beta` to the horizontal. Then it strikes the hill horizontally if

Options

A. `tanalpha= 2tanbeta`
B. `tanalpha= tanbeta`
C. `tanalpha. tanbeta= 1`
D. None

Answer

To solve the problem of determining the condition under which a shot fired at an angle \( \alpha \) to the horizontal strikes a hill inclined at an angle \( \beta \) horizontally, we can follow these steps:

### Step 1: Understand the Geometry of the Problem
- We have a projectile launched at an angle \( \alpha \) with respect to the horizontal.
- The hill is inclined at an angle \( \beta \) to the horizontal.
- When the projectile strikes the hill, it does so horizontally, meaning its vertical component of velocity at that point is zero.

### Step 2: Analyze the Components of Motion
- The initial velocity \( u \) can be broken down into its horizontal and vertical components:
- Horizontal component: \( u_x = u \cos \alpha \)
- Vertical component: \( u_y = u \sin \alpha \)

### Step 3: Determine the Condition for Horizontal Strike
- When the projectile strikes the hill horizontally, the vertical component of the velocity at the point of impact must be zero.
- The vertical motion can be described by the equation:
\[ v_y = u_y - g t \]
where \( g \) is the acceleration due to gravity and \( t \) is the time of flight.

- Setting \( v_y = 0 \) gives:
\[ 0 = u \sin \alpha - g t \implies t = \frac{u \sin \alpha}{g} \]

### Step 4: Calculate the Time of Flight to the Hill
- The time of flight until it strikes the hill can also be expressed in terms of the incline. The projectile travels a horizontal distance \( x \) and a vertical distance \( y \) until it reaches the hill.
- The vertical displacement of the projectile when it strikes the incline can be expressed in terms of the angle \( \beta \):
\[ y = x \tan \beta \]

### Step 5: Relate the Horizontal and Vertical Displacements
- The horizontal distance traveled in time \( t \) is:
\[ x = u \cos \alpha \cdot t = u \cos \alpha \cdot \frac{u \sin \alpha}{g} = \frac{u^2 \sin \alpha \cos \alpha}{g} \]

- The vertical displacement at the same time is:
\[ y = u \sin \alpha \cdot t - \frac{1}{2} g t^2 = u \sin \alpha \cdot \frac{u \sin \alpha}{g} - \frac{1}{2} g \left( \frac{u \sin \alpha}{g} \right)^2 \]
Simplifying gives:
\[ y = \frac{u^2 \sin^2 \alpha}{g} - \frac{1}{2} \frac{u^2 \sin^2 \alpha}{g} = \frac{u^2 \sin^2 \alpha}{2g} \]

### Step 6: Set the Vertical Displacement Equal to the Hill's Height
- Setting the vertical displacement equal to the height of the incline:
\[ \frac{u^2 \sin^2 \alpha}{2g} = \left( \frac{u^2 \sin \alpha \cos \beta}{g} \right) \tan \beta \]
Rearranging gives:
\[ \sin^2 \alpha = 2 \sin \alpha \cos \beta \tan \beta \]

### Step 7: Simplify to Find the Relation Between Angles
- This leads to the relation:
\[ \tan \alpha = 2 \tan \beta \]

### Final Conclusion
Thus, the condition for the projectile to strike the hill horizontally is:
\[ \tan \alpha = 2 \tan \beta \]