A particle is projected with velocity of `10m/s` at an angle of 15° with horizontal.The horizontal range will be`(g = 10m/s^2)`

Options

A. 10m
B. 5m
C. 2.5m
D. 1m

Answer

To find the horizontal range of a particle projected with an initial velocity at an angle, we can use the formula for the range \( R \) of projectile motion:

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

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

### Step 1: Identify the given values
- Initial velocity \( u = 10 \, \text{m/s} \)
- Angle of projection \( \theta = 15^\circ \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate \( \sin(2\theta) \)
First, we need to calculate \( 2\theta \):
\[ 2\theta = 2 \times 15^\circ = 30^\circ \]

Now, we find \( \sin(30^\circ) \):
\[ \sin(30^\circ) = \frac{1}{2} \]

### Step 3: Substitute the values into the range formula
Now we can substitute \( u \), \( \sin(2\theta) \), and \( g \) into the range formula:
\[ R = \frac{(10 \, \text{m/s})^2 \cdot \sin(30^\circ)}{10 \, \text{m/s}^2} \]

### Step 4: Calculate \( R \)
Calculating \( (10 \, \text{m/s})^2 \):
\[ (10 \, \text{m/s})^2 = 100 \, \text{m}^2/\text{s}^2 \]

Now substituting into the formula:
\[ R = \frac{100 \, \text{m}^2/\text{s}^2 \cdot \frac{1}{2}}{10 \, \text{m/s}^2} \]
\[ R = \frac{100 \cdot \frac{1}{2}}{10} = \frac{50}{10} = 5 \, \text{m} \]

### Final Answer
The horizontal range \( R \) is \( 5 \, \text{m} \).

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