A body is projected with a certain speed at angles of projection of `theta` and `90 - theta` The maximum height attained in the two cases are 20m and 10m respectively. The range for angle of projection `theta` is

Options

A. `40sqrt(2)` m
B. `20sqrt(2)`m
C. `80sqrt(2)` m
D. `200sqrt(2)`m

Answer

To solve the problem step by step, we will use the concepts of projectile motion, specifically focusing on the relationships between maximum height, angle of projection, and range.

### Step 1: Understanding the Maximum Height Formula
The maximum height \( H \) attained by a projectile is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

### Step 2: Setting Up the Equations
For the two angles of projection \( \theta \) and \( 90 - \theta \):
- For angle \( \theta \):
\[ H_1 = \frac{u^2 \sin^2 \theta}{2g} = 20 \text{ m} \]
- For angle \( 90 - \theta \):
\[ H_2 = \frac{u^2 \sin^2 (90 - \theta)}{2g} = \frac{u^2 \cos^2 \theta}{2g} = 10 \text{ m} \]

### Step 3: Forming the Ratio of Heights
Now, we can form the ratio of the two heights:
\[ \frac{H_1}{H_2} = \frac{\frac{u^2 \sin^2 \theta}{2g}}{\frac{u^2 \cos^2 \theta}{2g}} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta \]
Substituting the values of \( H_1 \) and \( H_2 \):
\[ \frac{20}{10} = \tan^2 \theta \implies \tan^2 \theta = 2 \]
Thus, we find:
\[ \tan \theta = \sqrt{2} \]

### Step 4: Finding Sine and Cosine Values
Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
Let \( \sin \theta = k \) and \( \cos \theta = \frac{k}{\sqrt{2}} \).
Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ k^2 + \left(\frac{k}{\sqrt{2}}\right)^2 = 1 \implies k^2 + \frac{k^2}{2} = 1 \implies \frac{3k^2}{2} = 1 \implies k^2 = \frac{2}{3} \implies k = \frac{1}{\sqrt{3}} \]
Thus,
\[ \sin \theta = \frac{1}{\sqrt{3}}, \quad \cos \theta = \frac{1}{\sqrt{2}} \]

### Step 5: Finding the Initial Velocity
Substituting \( \sin^2 \theta \) back into the height equation for \( H_1 \):
\[ 20 = \frac{u^2 \left(\frac{1}{\sqrt{3}}\right)^2}{2g} \implies 20 = \frac{u^2/3}{2g} \implies 20 = \frac{u^2}{6g} \implies u^2 = 120g \]

### Step 6: Calculating the Range
The range \( R \) for angle \( \theta \) is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Using \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ R = \frac{u^2 \cdot 2 \cdot \frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{2}}}{g} = \frac{u^2 \cdot \frac{2}{\sqrt{6}}}{g} \]
Substituting \( u^2 = 120g \):
\[ R = \frac{120g \cdot \frac{2}{\sqrt{6}}}{g} = 120 \cdot \frac{2}{\sqrt{6}} = 40\sqrt{6} \]

### Final Answer
Thus, the range for angle of projection \( \theta \) is:
\[ R = 40\sqrt{6} \text{ m} \]