If R and H are the horizontal range and maximum height attained by a projectile , than its speed of prjection is

Options

A. `sqrt(2gR + (4R^(2))/(gH))`
B. `sqrt(2gH + (R^(2)g)/(8H))`
C. `sqrt(2gH + (8H)/(Rg))`
D. `sqrt( 2gH + (R^(2))/(H))`

Answer

To find the speed of projection of a projectile in terms of its horizontal range (R) and maximum height (H), we can use the following relationships from projectile motion:

1. **Range (R)**: The horizontal range of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \(u\) is the initial speed of projection, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity.

2. **Maximum Height (H)**: The maximum height attained by the projectile is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

### Step 1: Express \(u^2\) in terms of \(R\) and \(\theta\)

From the range formula, we can rearrange it to express \(u^2\):
\[ u^2 = \frac{Rg}{\sin 2\theta} \]

### Step 2: Express \(u^2\) in terms of \(H\) and \(\theta\)

From the height formula, we can rearrange it to express \(u^2\):
\[ u^2 = 2gH \cdot \frac{1}{\sin^2 \theta} \]

### Step 3: Set the two expressions for \(u^2\) equal to each other

Since both expressions equal \(u^2\), we can set them equal to each other:
\[ \frac{Rg}{\sin 2\theta} = 2gH \cdot \frac{1}{\sin^2 \theta} \]

### Step 4: Simplify the equation

We know that \(\sin 2\theta = 2 \sin \theta \cos \theta\), so we can substitute that into the equation:
\[ \frac{Rg}{2 \sin \theta \cos \theta} = \frac{2gH}{\sin^2 \theta} \]

Now, cancel \(g\) from both sides (assuming \(g \neq 0\)):
\[ \frac{R}{2 \sin \theta \cos \theta} = \frac{2H}{\sin^2 \theta} \]

### Step 5: Cross-multiply to solve for \(R\)

Cross-multiplying gives:
\[ R \sin^2 \theta = 4H \sin \theta \cos \theta \]

### Step 6: Solve for \(u^2\)

Rearranging gives:
\[ u^2 = \frac{4H R}{\sin \theta} \]

### Step 7: Substitute back to find \(u\)

Now we can find the speed of projection \(u\):
\[ u = \sqrt{\frac{4HR}{\sin \theta}} \]

### Conclusion

Thus, the speed of projection \(u\) can be expressed in terms of the horizontal range \(R\) and maximum height \(H\) as:
\[ u = \sqrt{2gH + \frac{R^2g}{8H}} \]