A body is projected with a initial velocity of `(8hati+6hatj) m s^(-1)` . The horizontal range is `(g = 10 m s^(-2))`

Options

A. 9.6 m
B. 14 m
C. 50 m
D. 100 m

Answer

To find the horizontal range of a projectile launched with an initial velocity of \( \vec{u} = 8 \hat{i} + 6 \hat{j} \) m/s, we can follow these steps:

### Step 1: Identify the components of the initial velocity
The initial velocity can be broken down into its horizontal and vertical components:
- Horizontal component (\( u_x \)): \( 8 \) m/s
- Vertical component (\( u_y \)): \( 6 \) m/s

### Step 2: Calculate the time of flight
The time of flight (\( T \)) for a projectile can be calculated using the formula:
\[ T = \frac{2u_y}{g} \]
where \( g \) is the acceleration due to gravity. Given \( g = 10 \, \text{m/s}^2 \):
\[ T = \frac{2 \times 6}{10} = \frac{12}{10} = 1.2 \, \text{s} \]

### Step 3: Calculate the horizontal range
The horizontal range (\( R \)) can be calculated using the formula:
\[ R = u_x \times T \]
Substituting the values we have:
\[ R = 8 \, \text{m/s} \times 1.2 \, \text{s} = 9.6 \, \text{m} \]

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

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