A projectile aimed at a mark, which is in the horizontal plane through the point of projection, falls a cm short of it when the elevation is `alpha` and goes b cm far when the elevation is `beta`. Show that, if the speed of projection is same in all the cases the proper elevation is
`1/2 sin^(-1) [(bsin2alpha+asin2beta)/(a+b)]` .

Explanation

To solve the problem, we need to analyze the projectile motion of an object launched at different angles and determine the proper elevation angle that would allow it to hit a target. The projectile falls short of the target by a distance \( a \) when launched at an angle \( \alpha \) and goes beyond the target by a distance \( b \) when launched at an angle \( \beta \).

### Step-by-Step Solution:

1. **Understanding the Range of a Projectile**:
The range \( R \) of a projectile launched with an initial speed \( u \) at an angle \( \theta \) is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity.

2. **Setting Up the Equations**:
- When the projectile is launched at angle \( \alpha \), it falls short of the target by \( a \):
\[ R_1 = R - a = \frac{u^2 \sin(2\alpha)}{g} \]
- When the projectile is launched at angle \( \beta \), it goes beyond the target by \( b \):
\[ R_2 = R + b = \frac{u^2 \sin(2\beta)}{g} \]

3. **Expressing Ranges in Terms of R**:
From the above equations, we can express \( R \) in terms of \( R_1 \) and \( R_2 \):
\[ R = R_1 + a \quad \text{and} \quad R = R_2 - b \]

4. **Equating the Two Expressions for R**:
Setting the two expressions for \( R \) equal gives:
\[ R_1 + a = R_2 - b \]

5. **Substituting the Ranges**:
Substituting the expressions for \( R_1 \) and \( R_2 \):
\[ \frac{u^2 \sin(2\alpha)}{g} + a = \frac{u^2 \sin(2\beta)}{g} - b \]

6. **Rearranging the Equation**:
Rearranging the equation gives:
\[ \frac{u^2 \sin(2\beta)}{g} - \frac{u^2 \sin(2\alpha)}{g} = a + b \]

7. **Factoring Out Common Terms**:
Factoring out \( \frac{u^2}{g} \):
\[ \frac{u^2}{g} (\sin(2\beta) - \sin(2\alpha)) = a + b \]

8. **Finding the Proper Elevation**:
We want to find the angle \( \theta \) such that:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Using the previous results, we can express \( R \) in terms of \( a \) and \( b \):
\[ R = \frac{u^2}{g} \cdot \frac{b \sin(2\alpha) + a \sin(2\beta)}{a + b} \]

9. **Final Expression for the Angle**:
Setting the two expressions for \( R \) equal gives:
\[ \sin(2\theta) = \frac{b \sin(2\alpha) + a \sin(2\beta)}{a + b} \]
Therefore, the proper elevation angle \( \theta \) is:
\[ \theta = \frac{1}{2} \sin^{-1} \left( \frac{b \sin(2\alpha) + a \sin(2\beta)}{a + b} \right) \]

### Final Answer:
\[ \theta = \frac{1}{2} \sin^{-1} \left( \frac{b \sin(2\alpha) + a \sin(2\beta)}{a + b} \right) \]