Two balls are projected at an angle `theta` and `(90^(@) - theta)` to the horizontal with the same speed. The ratio of their maximum vertical heights is

Options

A. ` 1 : 1`
B. `tan theta : 1`
C. `1 : tan theta`
D. `tan^(2)theta : 1`

Explanation

To solve the problem, we need to find the ratio of the maximum vertical heights of two balls projected at angles \( \theta \) and \( 90^\circ - \theta \) with the same initial speed \( u \).

### Step-by-Step Solution:

1. **Identify 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 \( u \) is the initial speed, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of projection.

2. **Calculate the maximum height for the first ball**:
For the first ball projected at angle \( \theta \):
\[ h_1 = \frac{u^2 \sin^2 \theta}{2g} \]

3. **Calculate the maximum height for the second ball**:
For the second ball projected at angle \( 90^\circ - \theta \):
\[ h_2 = \frac{u^2 \sin^2 (90^\circ - \theta)}{2g} \]
Using the identity \( \sin(90^\circ - \theta) = \cos \theta \):
\[ h_2 = \frac{u^2 \cos^2 \theta}{2g} \]

4. **Find the ratio of the maximum heights**:
The ratio of the maximum heights \( \frac{h_1}{h_2} \) can be calculated as follows:
\[ \frac{h_1}{h_2} = \frac{\frac{u^2 \sin^2 \theta}{2g}}{\frac{u^2 \cos^2 \theta}{2g}} \]
Simplifying this expression:
\[ \frac{h_1}{h_2} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta \]

5. **State the final result**:
Therefore, the ratio of the maximum vertical heights of the two balls is:
\[ h_1 : h_2 = \tan^2 \theta : 1 \]

### Final Answer:
The ratio of their maximum vertical heights is \( \tan^2 \theta : 1 \).