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 of finding the ratio of the maximum vertical heights of two balls projected at angles θ and (90° - θ) with the same speed, we can follow these steps:

### Step 1: Understand the formula for maximum height in projectile motion
The maximum height \( h \) reached by a projectile launched with an initial speed \( u \) at an angle \( \theta \) is given by the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
where \( g \) is the acceleration due to gravity.

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

### Step 3: Calculate the maximum height for the second ball (angle 90° - θ)
For the second ball projected at angle \( 90° - \theta \):
\[ h_2 = \frac{u^2 \sin^2 (90° - \theta)}{2g} \]
Using the trigonometric identity \( \sin(90° - \theta) = \cos \theta \), we can rewrite this as:
\[ h_2 = \frac{u^2 \cos^2 \theta}{2g} \]

### Step 4: Find the ratio of the maximum heights
Now, we need to find the ratio \( \frac{h_1}{h_2} \):
\[ \frac{h_1}{h_2} = \frac{\frac{u^2 \sin^2 \theta}{2g}}{\frac{u^2 \cos^2 \theta}{2g}} \]
The \( u^2 \) and \( 2g \) terms cancel out:
\[ \frac{h_1}{h_2} = \frac{\sin^2 \theta}{\cos^2 \theta} \]

### Step 5: Simplify the ratio
Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ \frac{h_1}{h_2} = \tan^2 \theta \]

### Conclusion
Thus, the ratio of the maximum vertical heights of the two balls is:
\[ \frac{h_1}{h_2} = \tan^2 \theta \]