A bullet fired at an angle of ` 60^(@)` with the vertical hits the levelled ground at a distance of 200 m . Find the distance at which the bullet will hit the ground when fired at angle of ` 30^(@)`. (with same speed).

Answer

To solve the problem, we need to find the horizontal range of a bullet fired at an angle of \(30^\circ\) with the same initial speed as when it was fired at \(60^\circ\). We know that 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 speed,
- \(\theta\) is the angle of projection,
- \(g\) is the acceleration due to gravity.

### Step 1: Identify the given values
From the problem, we know:
- The range \(R\) when the bullet is fired at \(60^\circ\) is \(200 \, \text{m}\).
- The angle of projection for the first case, \(\theta = 60^\circ\).
- The angle of projection for the second case, \(\theta' = 30^\circ\).

### Step 2: Write the range formula for both angles
For the angle \(60^\circ\):
\[ R = \frac{u^2 \sin(2 \times 60^\circ)}{g} \]
\[ R = \frac{u^2 \sin(120^\circ)}{g} \]

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

### Step 3: Set up the ratio of the ranges
Since the speed \(u\) and \(g\) are the same in both cases, we can set up the ratio of the ranges:
\[ \frac{R'}{R} = \frac{\sin(60^\circ)}{\sin(120^\circ)} \]

### Step 4: Substitute the known values
We know that:
- \(R = 200 \, \text{m}\)
- \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\)
- \(\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}\)

Thus, substituting these values into the ratio:
\[ \frac{R'}{200} = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} = 1 \]

### Step 5: Solve for \(R'\)
From the ratio, we find:
\[ R' = 200 \, \text{m} \]

### Conclusion
The distance at which the bullet will hit the ground when fired at an angle of \(30^\circ\) is also \(200 \, \text{m}\).