Two balls are projected with the same velocity but with different angles with the horizontal. Their ranges are equal. If the angle of projection of one is 30° and its maximum height is h, then the maximum height of other will be

Options

A. 1h
B. 3h
C. 6h
D. 10h

Explanation

To solve the problem, we need to find the maximum height of the second ball projected at an angle of 60° given that the first ball is projected at an angle of 30° and its maximum height is \( h \).

### Step-by-Step Solution:

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

2. **Calculate Maximum Height for the First Ball**:
For the first ball projected at an angle of \( 30^\circ \):
\[ h = \frac{u^2 \sin^2(30^\circ)}{2g} \]
We know that \( \sin(30^\circ) = \frac{1}{2} \), so:
\[ h = \frac{u^2 \left(\frac{1}{2}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{4}}{2g} = \frac{u^2}{8g} \]

3. **Calculate Maximum Height for the Second Ball**:
For the second ball projected at an angle of \( 60^\circ \):
\[ h' = \frac{u^2 \sin^2(60^\circ)}{2g} \]
We know that \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \), so:
\[ h' = \frac{u^2 \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{u^2 \cdot \frac{3}{4}}{2g} = \frac{3u^2}{8g} \]

4. **Finding the Ratio of Heights**:
Now, we can find the ratio of the maximum heights \( h \) and \( h' \):
\[ \frac{h'}{h} = \frac{\frac{3u^2}{8g}}{\frac{u^2}{8g}} = 3 \]
This implies:
\[ h' = 3h \]

### Conclusion:
Thus, the maximum height of the second ball, which is projected at \( 60^\circ \), is \( 3h \).