A projectile is projected with the initial velocity `(6i + 8j) m//s`. The horizontal range is `(g = 10 m//s^(2))`

Options

A. `96 m`
B. `960 m`
C. `9.6 m`
D. `4.8 m`

Explanation

To solve the problem of finding the horizontal range of a projectile projected with an initial velocity of \( (6i + 8j) \, m/s \), we can follow these steps:

### Step 1: Identify the components of the initial velocity
The initial velocity vector is given as:
\[ \vec{u} = 6\hat{i} + 8\hat{j} \, m/s \]
From this, we can identify:
- The horizontal component \( u_x = 6 \, m/s \)
- The vertical component \( u_y = 8 \, m/s \)

### Step 2: Calculate the magnitude of the initial velocity
The magnitude of the initial velocity \( u \) can be calculated using the Pythagorean theorem:
\[ u = \sqrt{u_x^2 + u_y^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, m/s \]

### Step 3: Calculate the angle of projection
The angle of projection \( \theta \) can be found using the trigonometric ratios:
\[ \tan \theta = \frac{u_y}{u_x} = \frac{8}{6} \]
To find \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{8}{6}\right) = \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 4: Use the range formula
The formula for the horizontal range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Where \( g = 10 \, m/s^2 \). We need to calculate \( \sin 2\theta \):
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
From the components, we can find:
\[ \sin \theta = \frac{u_y}{u} = \frac{8}{10} = 0.8 \]
\[ \cos \theta = \frac{u_x}{u} = \frac{6}{10} = 0.6 \]
Now, substituting these values:
\[ \sin 2\theta = 2 \cdot 0.8 \cdot 0.6 = 0.96 \]

### Step 5: Substitute values into the range formula
Now we can substitute \( u \), \( \sin 2\theta \), and \( g \) into the range formula:
\[ R = \frac{10^2 \cdot 0.96}{10} = \frac{100 \cdot 0.96}{10} = 9.6 \, m \]

### Conclusion
The horizontal range of the projectile is:
\[ \boxed{9.6 \, m} \]

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