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 \( u \) of a projectile in terms of its horizontal range \( R \) and maximum height \( H \), we can use the equations of projectile motion. Here’s a step-by-step solution:

### Step 1: Understand the equations of motion
In projectile motion, the horizontal range \( R \) and the maximum height \( H \) can be expressed in terms of the initial speed \( u \) and the angle of projection \( \theta \).

### Step 2: Write the equations for maximum height and range
1. The maximum height \( H \) is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( g \) is the acceleration due to gravity.

2. The horizontal range \( R \) is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we can rewrite it as:
\[ R = \frac{u^2 \cdot 2 \sin \theta \cos \theta}{g} \]

### Step 3: Express \( \sin \theta \) and \( \cos \theta \) in terms of \( H \) and \( R \)
From the equation for maximum height, we can express \( \sin^2 \theta \):
\[ \sin^2 \theta = \frac{2gH}{u^2} \]
Thus, we can express \( \sin \theta \) as:
\[ \sin \theta = \sqrt{\frac{2gH}{u^2}} \]

Now, substituting \( \sin \theta \) into the range equation:
\[ R = \frac{u^2 \cdot 2 \sqrt{\frac{2gH}{u^2}} \cdot \cos \theta}{g} \]

### Step 4: Solve for \( \cos \theta \)
From the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{2gH}{u^2} \]
Thus, \( \cos \theta = \sqrt{1 - \frac{2gH}{u^2}} \).

### Step 5: Substitute \( \cos \theta \) back into the range equation
Now substituting \( \cos \theta \) into the range equation:
\[ R = \frac{u^2 \cdot 2 \sqrt{\frac{2gH}{u^2}} \cdot \sqrt{1 - \frac{2gH}{u^2}}}{g} \]

### Step 6: Simplify the equation
Squaring both sides and simplifying will yield:
\[ R^2 = \frac{4u^2 \cdot 2gH \cdot (1 - \frac{2gH}{u^2})}{g^2} \]
This can be rearranged to isolate \( u^2 \).

### Step 7: Final expression for \( u \)
After simplifying, we can express \( u \) in terms of \( R \) and \( H \):
\[ u^2 = \frac{gR^2}{8H} \]
Thus,
\[ u = \sqrt{\frac{gR^2}{8H}} + \sqrt{2gH} \]

### Final Result
The speed of projection \( u \) is given by:
\[ u = \sqrt{2gH + \frac{gR^2}{8H}} \]