Two bodies are thrown up at angles of `45^(@)` and `60^(@)`, respectively, with the horizontal. If both bodies attain same vertical height, then the ratio of velocities with which these are thrown is

Options

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

Answer

To solve the problem, we need to use the formula for the maximum height attained by a projectile. The maximum height \( H \) reached by a projectile thrown with an initial velocity \( u \) at an angle \( \theta \) is given by:

\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

where \( g \) is the acceleration due to gravity.

### Step 1: Write the height equations for both projectiles

Let \( u_1 \) be the initial velocity of the first body thrown at an angle of \( 45^\circ \) and \( u_2 \) be the initial velocity of the second body thrown at an angle of \( 60^\circ \).

For the first body:
\[ H_1 = \frac{u_1^2 \sin^2(45^\circ)}{2g} \]

For the second body:
\[ H_2 = \frac{u_2^2 \sin^2(60^\circ)}{2g} \]

### Step 2: Substitute the values of \( \sin(45^\circ) \) and \( \sin(60^\circ) \)

We know:
\[ \sin(45^\circ) = \frac{\sqrt{2}}{2} \quad \text{and} \quad \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

Substituting these values into the height equations:

For the first body:
\[ H_1 = \frac{u_1^2 \left(\frac{\sqrt{2}}{2}\right)^2}{2g} = \frac{u_1^2 \cdot \frac{1}{2}}{2g} = \frac{u_1^2}{4g} \]

For the second body:
\[ H_2 = \frac{u_2^2 \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{u_2^2 \cdot \frac{3}{4}}{2g} = \frac{3u_2^2}{8g} \]

### Step 3: Set the heights equal to each other

Since both bodies attain the same vertical height, we can set \( H_1 = H_2 \):

\[ \frac{u_1^2}{4g} = \frac{3u_2^2}{8g} \]

### Step 4: Simplify the equation

We can cancel \( g \) from both sides:

\[ \frac{u_1^2}{4} = \frac{3u_2^2}{8} \]

Now, cross-multiply to eliminate the fractions:

\[ 8u_1^2 = 12u_2^2 \]

### Step 5: Solve for the ratio \( \frac{u_1}{u_2} \)

Rearranging gives:

\[ \frac{u_1^2}{u_2^2} = \frac{12}{8} = \frac{3}{2} \]

Taking the square root of both sides:

\[ \frac{u_1}{u_2} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2} \]

### Final Answer

Thus, the ratio of the velocities with which the two bodies are thrown is:

\[ \frac{u_1}{u_2} = \frac{\sqrt{6}}{2} \]