The velocity of projection of oblique projectile is `(6hati+8hatj)ms^(-1)` The horizontal range of the projectile is

Options

A. 4.8 metre
B. 9.6 metre
C. 19.2 metre
D. 14.0 metre

Explanation

To find the horizontal range of the oblique projectile, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Components of Velocity**:
The given velocity of projection is \( \vec{u} = 6 \hat{i} + 8 \hat{j} \) m/s.
- The horizontal component of velocity \( u_x = 6 \) m/s.
- The vertical component of velocity \( u_y = 8 \) m/s.

2. **Determine the Magnitude of Initial Velocity**:
The magnitude of the initial velocity \( u \) can be calculated using the Pythagorean theorem:
\[ u = \sqrt{u_x^2 + u_y^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ m/s} \]

3. **Calculate the Angle of Projection**:
The angle \( \theta \) of projection can be found using the tangent function:
\[ \tan \theta = \frac{u_y}{u_x} = \frac{8}{6} = \frac{4}{3} \]
Thus, \( \theta = \tan^{-1}\left(\frac{4}{3}\right) \).

4. **Use the Range Formula**:
The horizontal 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 \( 10 \text{ m/s}^2 \)).

5. **Calculate \( \sin 2\theta \)**:
We know that:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
To find \( \sin \theta \) and \( \cos \theta \):
\[ \sin \theta = \frac{u_y}{u} = \frac{8}{10} = 0.8 \]
\[ \cos \theta = \frac{u_x}{u} = \frac{6}{10} = 0.6 \]
Therefore,
\[ \sin 2\theta = 2 \cdot 0.8 \cdot 0.6 = 0.96 \]

6. **Substitute Values into the Range Formula**:
Now substitute \( u = 10 \text{ m/s} \), \( \sin 2\theta = 0.96 \), and \( g = 10 \text{ m/s}^2 \) into the range formula:
\[ R = \frac{10^2 \cdot 0.96}{10} = \frac{100 \cdot 0.96}{10} = 9.6 \text{ m} \]

### Final Answer
The horizontal range of the projectile is \( 9.6 \text{ m} \).

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