A body is prodected with initial velocity` ( 5hati + 12hati)` m//s from origin . Gravity acts in negative y direction. The horizontal rang is (Take g = 10 `m//s^(2)`)

Options

A. 12m
B. 18 m
C. 17 m
D. 7 m

Answer

To solve the problem of finding the horizontal range of a body projected with an initial velocity of \( \vec{u} = 5 \hat{i} + 12 \hat{j} \) m/s from the origin, 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:
- \( u_x = 5 \) m/s (horizontal component)
- \( u_y = 12 \) m/s (vertical component)

### Step 2: Calculate the time of flight
The time of flight \( T \) can be calculated using the formula:
\[ T = \frac{2u_y}{g} \]
where \( g \) is the acceleration due to gravity. Given \( g = 10 \, \text{m/s}^2 \):
\[ T = \frac{2 \times 12}{10} = \frac{24}{10} = 2.4 \, \text{s} \]

### 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:
\[ R = 5 \times 2.4 = 12 \, \text{m} \]

### Conclusion
The horizontal range of the body is \( 12 \, \text{m} \).

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