If 2 balls are projected at angles `45^(@)` and `60^(@)` and the maximum heights reached are same, what is the ratio of their initial velocities ?

Options

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

Explanation

To find the ratio of the initial velocities of two balls projected at angles of 45 degrees and 60 degrees, given that they reach the same maximum height, we can follow these steps:

### Step 1: Understand the maximum height formula
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 velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

### Step 2: Set up the equations for both balls
Let:
- For ball A (projected at \( 45^\circ \)):
\[ H_A = \frac{v_1^2 \sin^2 45^\circ}{2g} \]
- For ball B (projected at \( 60^\circ \)):
\[ H_B = \frac{v_2^2 \sin^2 60^\circ}{2g} \]

Since both balls reach the same maximum height, we can set \( H_A = H_B \):
\[ \frac{v_1^2 \sin^2 45^\circ}{2g} = \frac{v_2^2 \sin^2 60^\circ}{2g} \]

### Step 3: Cancel common terms
The \( 2g \) in the denominator cancels out:
\[ v_1^2 \sin^2 45^\circ = v_2^2 \sin^2 60^\circ \]

### Step 4: Substitute the sine values
We know:
- \( \sin 45^\circ = \frac{1}{\sqrt{2}} \)
- \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)

Substituting these values into the equation gives:
\[ v_1^2 \left(\frac{1}{\sqrt{2}}\right)^2 = v_2^2 \left(\frac{\sqrt{3}}{2}\right)^2 \]
This simplifies to:
\[ v_1^2 \cdot \frac{1}{2} = v_2^2 \cdot \frac{3}{4} \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ \frac{v_1^2}{v_2^2} = \frac{3/4}{1/2} \]
This simplifies to:
\[ \frac{v_1^2}{v_2^2} = \frac{3}{2} \]

### Step 6: Taking the square root
Taking the square root of both sides gives:
\[ \frac{v_1}{v_2} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} \]

### Step 7: Conclusion
Thus, the ratio of their initial velocities is:
\[ v_1 : v_2 = \sqrt{3} : \sqrt{2} \]