Two bodies are thrown with the same initial velocity at angles ` theta ` and `(90^@ - theta)` respectively with the horizontal, then their maximum height are in the ratio

Options

A. 1 : 1
B. ` sintheta : costheta`
C. `sin^2theta:cos^2theta`
D. `costheta:sintheta`

Explanation

To solve the problem of finding the ratio of maximum heights of two bodies thrown at angles \( \theta \) and \( 90^\circ - \theta \) with the same initial velocity, we can follow these steps:

### Step 1: Write the 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:
- \( u \) = initial velocity
- \( g \) = acceleration due to gravity
- \( \theta \) = angle of projection

### Step 2: Calculate the maximum height for the first angle \( \theta \)
For the first body thrown at angle \( \theta \):
\[ H_1 = \frac{u^2 \sin^2 \theta}{2g} \]

### Step 3: Calculate the maximum height for the second angle \( 90^\circ - \theta \)
For the second body thrown at angle \( 90^\circ - \theta \):
\[ H_2 = \frac{u^2 \sin^2 (90^\circ - \theta)}{2g} \]
Using the trigonometric identity, we know that:
\[ \sin(90^\circ - \theta) = \cos \theta \]
Thus, we can rewrite \( H_2 \) as:
\[ H_2 = \frac{u^2 \cos^2 \theta}{2g} \]

### Step 4: Find the ratio of the maximum heights
Now we can find the ratio of the maximum heights \( H_1 \) and \( H_2 \):
\[ \text{Ratio} = \frac{H_1}{H_2} = \frac{\frac{u^2 \sin^2 \theta}{2g}}{\frac{u^2 \cos^2 \theta}{2g}} \]
The \( u^2 \) and \( 2g \) terms cancel out:
\[ \text{Ratio} = \frac{\sin^2 \theta}{\cos^2 \theta} \]

### Step 5: Simplify the ratio
This can be further simplified to:
\[ \text{Ratio} = \tan^2 \theta \]

### Conclusion
Thus, the ratio of the maximum heights of the two bodies thrown at angles \( \theta \) and \( 90^\circ - \theta \) is:
\[ \text{Ratio} = \tan^2 \theta \]