Two particles are projected with same velocity but at angles of projection 25° and 65° with horizontal. The ratio of their horizontal ranges is

Options

A. `1:2`
B. `1:1`
C. `2:1`
D. cannot be determined

Answer

To find the ratio of the horizontal ranges of two particles projected at different angles but with the same initial velocity, we can use the formula for the horizontal range of a projectile:

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

where:
- \( R \) is the range,
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

### Step 1: Write the expressions for the ranges of both particles.

Let:
- \( R_1 \) be the range of the first particle projected at an angle of \( 25^\circ \).
- \( R_2 \) be the range of the second particle projected at an angle of \( 65^\circ \).

Using the range formula, we have:
\[ R_1 = \frac{u^2 \sin(2 \times 25^\circ)}{g} \]
\[ R_2 = \frac{u^2 \sin(2 \times 65^\circ)}{g} \]

### Step 2: Calculate \( \sin(2 \theta) \) for both angles.

Calculating \( \sin(2 \times 25^\circ) \):
\[ \sin(50^\circ) \]

Calculating \( \sin(2 \times 65^\circ) \):
\[ \sin(130^\circ) \]

### Step 3: Use the property of sine function.

We know that:
\[ \sin(130^\circ) = \sin(180^\circ - 50^\circ) = \sin(50^\circ) \]

### Step 4: Set up the ratio of the ranges.

Now, we can find the ratio of the ranges:
\[ \frac{R_1}{R_2} = \frac{\frac{u^2 \sin(50^\circ)}{g}}{\frac{u^2 \sin(130^\circ)}{g}} = \frac{\sin(50^\circ)}{\sin(50^\circ)} = 1 \]

### Conclusion

Thus, the ratio of their horizontal ranges is:
\[ \frac{R_1}{R_2} = 1:1 \]