Two balls are projected respectively from the same point in directions inclined at `60^(@)` and `30^(@)` to the horizontal. If they attain the same height, the ratio of their velocities of projection is

Options

A. `sqrt(3):1`
B. `1: sqrt(3)`
C. `1:1`
D. `1:2`

Answer

To solve the problem, we need to find the ratio of the velocities of projection of two balls projected at angles of \(60^\circ\) and \(30^\circ\) to the horizontal, given that they reach the same maximum height.

### Step-by-Step Solution:

1. **Understanding Maximum Height Formula**:
The maximum height \(H\) attained by 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. **Setting Up the Equations**:
For the first ball projected at \(60^\circ\):
\[ H_1 = \frac{u_1^2 \sin^2(60^\circ)}{2g} \]
For the second ball projected at \(30^\circ\):
\[ H_2 = \frac{u_2^2 \sin^2(30^\circ)}{2g} \]

3. **Equating the Heights**:
Since both balls attain the same height:
\[ H_1 = H_2 \]
Therefore,
\[ \frac{u_1^2 \sin^2(60^\circ)}{2g} = \frac{u_2^2 \sin^2(30^\circ)}{2g} \]
The \(2g\) cancels out from both sides:
\[ u_1^2 \sin^2(60^\circ) = u_2^2 \sin^2(30^\circ) \]

4. **Substituting Values for Sine**:
We know:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin(30^\circ) = \frac{1}{2} \]
Substituting these values into the equation:
\[ u_1^2 \left(\frac{\sqrt{3}}{2}\right)^2 = u_2^2 \left(\frac{1}{2}\right)^2 \]
Simplifying this gives:
\[ u_1^2 \cdot \frac{3}{4} = u_2^2 \cdot \frac{1}{4} \]

5. **Rearranging the Equation**:
Multiplying both sides by 4:
\[ 3u_1^2 = u_2^2 \]
Taking the square root of both sides:
\[ \frac{u_1}{u_2} = \frac{1}{\sqrt{3}} \]

6. **Finding the Ratio**:
Therefore, the ratio of their velocities of projection is:
\[ \frac{u_1}{u_2} = \frac{1}{\sqrt{3}} \quad \text{or} \quad u_1 : u_2 = 1 : \sqrt{3} \]

### Final Answer:
The ratio of their velocities of projection is \(1 : \sqrt{3}\).