A body is projected into air with a velocity (3i + 4j + 5k)m/s. Taking x-y plane as base and Z axis as vertical, the range of projectile (g =10`m//s^(2)`)

Options

A. 10 m, `5pi` m
B. 5 m
C. 20 m
D. 15 m

Answer

To solve the problem of finding the range of a projectile projected with a velocity of \( \mathbf{u} = (3\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}) \, \text{m/s} \), we will follow these steps:

### Step 1: Identify the components of the initial velocity
The initial velocity can be broken down into its components:
- \( u_x = 3 \, \text{m/s} \) (horizontal component in the x-direction)
- \( u_y = 4 \, \text{m/s} \) (horizontal component in the y-direction)
- \( u_z = 5 \, \text{m/s} \) (vertical component in the z-direction)

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

Substituting the values:
\[ T = \frac{2 \times 5}{10} = 1 \, \text{s} \]

### Step 3: Calculate the horizontal range
The horizontal range \( R \) can be calculated using the horizontal components of the velocity and the time of flight:
\[ R = (u_x + u_y) \times T \]
Here, we consider the combined horizontal velocity in the x-y plane:
\[ u_{horizontal} = \sqrt{u_x^2 + u_y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s} \]
Now substituting into the range equation:
\[ R = u_{horizontal} \times T = 5 \times 1 = 5 \, \text{m} \]

### Final Answer:
The range of the projectile is \( 5 \, \text{m} \).

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