In previous question, what is the relation between `h_1 and h_2`

Options

A. `h_1 (h_2)/4`
B. `h_1 (h_2)/3`
C. `h_1 h_2/2`
D. `h_1 =h_2`

Explanation

To find the relationship between the maximum heights \( h_1 \) and \( h_2 \) of two projectiles launched at different angles, we can follow these steps:

### Step 1: Understand the Angles of Projection
- The first projectile is launched at an angle of \( 30^\circ \) with the horizontal.
- The second projectile is launched at an angle of \( 60^\circ \) with the horizontal (which is complementary to \( 30^\circ \)).

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

### Step 3: Calculate Maximum Height \( h_1 \) for the First Projectile
For the first projectile launched at \( 30^\circ \):
\[ h_1 = \frac{u^2 \sin^2(30^\circ)}{2g} \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ h_1 = \frac{u^2 \left(\frac{1}{2}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{4}}{2g} = \frac{u^2}{8g} \]

### Step 4: Calculate Maximum Height \( h_2 \) for the Second Projectile
For the second projectile launched at \( 60^\circ \):
\[ h_2 = \frac{u^2 \sin^2(60^\circ)}{2g} \]
Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[ h_2 = \frac{u^2 \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{u^2 \cdot \frac{3}{4}}{2g} = \frac{3u^2}{8g} \]

### Step 5: Find the Relationship Between \( h_1 \) and \( h_2 \)
Now we have:
\[ h_1 = \frac{u^2}{8g} \quad \text{and} \quad h_2 = \frac{3u^2}{8g} \]
To find the relationship:
\[ \frac{h_1}{h_2} = \frac{\frac{u^2}{8g}}{\frac{3u^2}{8g}} = \frac{1}{3} \]
Thus, we can express this as:
\[ h_1 = \frac{h_2}{3} \]

### Conclusion
The relationship between the maximum heights \( h_1 \) and \( h_2 \) is:
\[ h_1 = \frac{h_2}{3} \]