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 solve the problem of finding the horizontal range of a particle projected with a velocity of 10 m/s at an angle of 15° with the horizontal, we can follow these steps:

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

### Step 2: Resolve the initial velocity into horizontal and vertical components
- The horizontal component of the velocity, \( u_x = u \cos \theta \)
- The vertical component of the velocity, \( u_y = u \sin \theta \)

Calculating these components:
- \( u_x = 10 \cos(15°) \)
- \( u_y = 10 \sin(15°) \)

Using approximate values:
- \( \cos(15°) \approx 0.9659 \)
- \( \sin(15°) \approx 0.2588 \)

Calculating:
- \( u_x \approx 10 \times 0.9659 \approx 9.659 \, \text{m/s} \)
- \( u_y \approx 10 \times 0.2588 \approx 2.588 \, \text{m/s} \)

### Step 3: Calculate the time of flight
The time of flight \( T \) can be calculated using the formula:
\[ T = \frac{2u_y}{g} \]
Substituting the values:
\[ T = \frac{2 \times 2.588}{10} \approx \frac{5.176}{10} \approx 0.5176 \, \text{s} \]

### Step 4: Calculate the horizontal range
The horizontal range \( R \) is given by:
\[ R = u_x \times T \]
Substituting the values:
\[ R = 9.659 \times 0.5176 \approx 5.000 \, \text{m} \]

### Conclusion
The horizontal range of the projectile is approximately \( 5 \, \text{m} \).