Two projectiles thrown from the same point at angles `60^@ and 30^@` with the horizontal attain the same height. The ratio of their initial velocities is

Options

A. (a)1
B. (b)2
C. (c)`sqrt(3)`
D. (d)`1/sqrt3`

Answer

To solve the problem of finding the ratio of the initial velocities of two projectiles thrown at angles of \(60^\circ\) and \(30^\circ\) that attain the same height, we can follow these steps:

### Step 1: Understand the formula for maximum height
The maximum height (\(h_{max}\)) reached by a projectile is given by the formula:
\[ h_{max} = \frac{u^2 \sin^2 \theta}{2g} \]
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 projectiles
Let:
- \(u_1\) be the initial velocity of the projectile thrown at \(60^\circ\)
- \(u_2\) be the initial velocity of the projectile thrown at \(30^\circ\)

Since both projectiles reach the same maximum height, we can equate their height equations:
\[ \frac{u_1^2 \sin^2(60^\circ)}{2g} = \frac{u_2^2 \sin^2(30^\circ)}{2g} \]

### Step 3: Simplify the equation
We can cancel \(2g\) from both sides:
\[ u_1^2 \sin^2(60^\circ) = u_2^2 \sin^2(30^\circ) \]

### Step 4: Solve for the ratio of initial velocities
Rearranging the equation gives:
\[ \frac{u_1^2}{u_2^2} = \frac{\sin^2(30^\circ)}{\sin^2(60^\circ)} \]
Taking the square root of both sides, we find:
\[ \frac{u_1}{u_2} = \frac{\sin(30^\circ)}{\sin(60^\circ)} \]

### Step 5: Substitute the known sine values
We know:
- \(\sin(30^\circ) = \frac{1}{2}\)
- \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\)

Substituting these values into the equation gives:
\[ \frac{u_1}{u_2} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \]

### Conclusion
Thus, the ratio of their initial velocities is:
\[ \frac{u_1}{u_2} = \frac{1}{\sqrt{3}} \]