Two bodies are thrown from the same point with the same velocity of `50ms^(-1)`.if their angles of projection are complimentary to each other and the difference of maximum heights is `30m`,the minimum and maximum heights are`(g=10 m//s^(2))`

Options

A. `50 m & 80m`
B. `47.5 m & 77.5 m`
C. `30 m & 60m`
D. `25 m & 55m`

Explanation

To solve the problem, we need to find the maximum and minimum heights of two bodies thrown with the same velocity at complementary angles. We know that the difference in their maximum heights is 30 m.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Two bodies are thrown with the same initial velocity \( u = 50 \, \text{m/s} \).
- The angles of projection are complementary, meaning if one angle is \( \theta \), the other angle is \( 90^\circ - \theta \).
- The difference in their maximum heights is given as \( h_2 - h_1 = 30 \, \text{m} \).

2. **Formula for Maximum Height**:
The maximum height \( h \) reached by a projectile is given by the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
where \( g \) is the acceleration due to gravity.

3. **Calculate Maximum Heights**:
- For the first body (angle \( \theta \)):
\[ h_1 = \frac{u^2 \sin^2 \theta}{2g} \]
- For the second body (angle \( 90^\circ - \theta \)):
\[ h_2 = \frac{u^2 \sin^2 (90^\circ - \theta)}{2g} = \frac{u^2 \cos^2 \theta}{2g} \]

4. **Setting Up the Equation**:
From the problem, we have:
\[ h_2 - h_1 = 30 \]
Substituting the expressions for \( h_1 \) and \( h_2 \):
\[ \frac{u^2 \cos^2 \theta}{2g} - \frac{u^2 \sin^2 \theta}{2g} = 30 \]
This simplifies to:
\[ \frac{u^2}{2g} (\cos^2 \theta - \sin^2 \theta) = 30 \]

5. **Using Trigonometric Identity**:
We know that \( \cos^2 \theta - \sin^2 \theta = \cos 2\theta \). Thus, we can rewrite the equation as:
\[ \frac{u^2}{2g} \cos 2\theta = 30 \]

6. **Substituting Values**:
Substitute \( u = 50 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ \frac{50^2}{2 \times 10} \cos 2\theta = 30 \]
Simplifying gives:
\[ \frac{2500}{20} \cos 2\theta = 30 \implies 125 \cos 2\theta = 30 \implies \cos 2\theta = \frac{30}{125} = \frac{6}{25} \]

7. **Finding Maximum Heights**:
Now we can find \( h_1 \) and \( h_2 \):
- From the earlier expression:
\[ h_1 + h_2 = \frac{u^2}{2g} = \frac{2500}{20} = 125 \, \text{m} \]
- We have two equations:
1. \( h_2 - h_1 = 30 \)
2. \( h_1 + h_2 = 125 \)

8. **Solving the Equations**:
Adding the two equations:
\[ (h_2 - h_1) + (h_1 + h_2) = 30 + 125 \]
\[ 2h_2 = 155 \implies h_2 = \frac{155}{2} = 77.5 \, \text{m} \]
Now substituting back to find \( h_1 \):
\[ h_1 = h_2 - 30 = 77.5 - 30 = 47.5 \, \text{m} \]

### Final Answer:
- Minimum Height \( h_1 = 47.5 \, \text{m} \)
- Maximum Height \( h_2 = 77.5 \, \text{m} \)