Two bodies of same mass are projected with the same velocity at an angle `30^(@)` and `60^(@)` respectively.The ration of their horizontal ranges will be

Options

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

Answer

To solve the problem of finding the ratio of the horizontal ranges of two bodies projected at angles of \(30^\circ\) and \(60^\circ\) with the same initial velocity, we can follow these steps:

### Step 1: Understand the formula for horizontal range
The horizontal range \(R\) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where:
- \(u\) is the initial velocity,
- \(\theta\) is the angle of projection,
- \(g\) is the acceleration due to gravity.

### Step 2: Calculate the range for both angles
Let’s denote the ranges for the angles \(30^\circ\) and \(60^\circ\) as \(R_1\) and \(R_2\) respectively.

For the angle \(30^\circ\):
\[ R_1 = \frac{u^2 \sin(2 \times 30^\circ)}{g} = \frac{u^2 \sin(60^\circ)}{g} \]

For the angle \(60^\circ\):
\[ R_2 = \frac{u^2 \sin(2 \times 60^\circ)}{g} = \frac{u^2 \sin(120^\circ)}{g} \]

### Step 3: Simplify the ratio of ranges
Now, we need to find the ratio \( \frac{R_1}{R_2} \):
\[ \frac{R_1}{R_2} = \frac{\frac{u^2 \sin(60^\circ)}{g}}{\frac{u^2 \sin(120^\circ)}{g}} = \frac{\sin(60^\circ)}{\sin(120^\circ)} \]

### Step 4: Use the sine values
We know that:
- \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\)
- \(\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}\)

Thus, substituting the sine values:
\[ \frac{R_1}{R_2} = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} = 1 \]

### Conclusion
The ratio of the horizontal ranges of the two bodies is:
\[ R_1 : R_2 = 1 : 1 \]