A particle is projected from ground with velocity `3hati + 4hatj` m/s. Find range of the projectile :-

Options

A. `1.2 m`
B. `3.6 m`
C. `2.4 m`
D. `10 m`

Answer

To find the range of a projectile projected from the ground with a velocity of \( \vec{u} = 3 \hat{i} + 4 \hat{j} \) m/s, we can follow these steps:

### Step 1: Identify the components of the initial velocity
The initial velocity \( \vec{u} \) can be broken down into its horizontal and vertical components:
- Horizontal component \( u_x = 3 \) m/s
- Vertical component \( u_y = 4 \) m/s

### Step 2: Calculate the magnitude of the initial velocity
The magnitude of the initial velocity \( u \) can be calculated using the Pythagorean theorem:
\[ u = \sqrt{u_x^2 + u_y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ m/s} \]

### Step 3: Determine the angle of projection
The angle \( \theta \) of projection can be found using the tangent function:
\[ \tan(\theta) = \frac{u_y}{u_x} = \frac{4}{3} \]
Thus, \( \theta = \tan^{-1}\left(\frac{4}{3}\right) \).

### Step 4: Use the range formula for projectile motion
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 5: Calculate \( \sin(2\theta) \)
Using the double angle identity:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
We can find \( \sin(\theta) \) and \( \cos(\theta) \):
- \( \sin(\theta) = \frac{u_y}{u} = \frac{4}{5} \)
- \( \cos(\theta) = \frac{u_x}{u} = \frac{3}{5} \)

Thus,
\[ \sin(2\theta) = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25} \]

### Step 6: Substitute values into the range formula
Now substituting the values into the range formula:
\[ R = \frac{(5)^2 \cdot \frac{24}{25}}{10} = \frac{25 \cdot \frac{24}{25}}{10} = \frac{24}{10} = 2.4 \text{ m} \]

### Conclusion
The range of the projectile is \( 2.4 \) meters.

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