A body is projected from the ground with a velocity `v = (3hati +10 hatj)ms^(-1)`. The maximum height attained and the range of the body respectively are (given `g = 10 ms^(-2))`

Options

A. 5 m and 6m
B. 3m and 10m
C. 6m and 5m
D. 3m and 5m

Explanation

To solve the problem of finding the maximum height and the range of a body projected with a velocity \( \mathbf{v} = (3 \hat{i} + 10 \hat{j}) \, \text{m/s} \) and given \( g = 10 \, \text{m/s}^2 \), we can follow these steps:

### Step 1: Identify the components of the initial velocity
The initial velocity vector can be broken down into its horizontal and vertical components:
- \( v_x = 3 \, \text{m/s} \) (horizontal component)
- \( v_y = 10 \, \text{m/s} \) (vertical component)

### Step 2: Calculate the maximum height
The formula for the maximum height \( h \) attained by a projectile is given by:
\[ h = \frac{v_y^2}{2g} \]
Substituting the known values:
\[ h = \frac{(10)^2}{2 \times 10} = \frac{100}{20} = 5 \, \text{m} \]

### Step 3: Calculate the range
The formula for the range \( R \) of a projectile is given by:
\[ R = \frac{2 v_x v_y}{g} \]
Substituting the known values:
\[ R = \frac{2 \times 3 \times 10}{10} = \frac{60}{10} = 6 \, \text{m} \]

### Final Result
The maximum height attained by the body is \( 5 \, \text{m} \) and the range of the body is \( 6 \, \text{m} \).

### Summary
- Maximum Height: \( 5 \, \text{m} \)
- Range: \( 6 \, \text{m} \)