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 a body projected from the ground with a velocity \( \mathbf{v} = (3 \hat{i} + 10 \hat{j}) \, \text{m/s} \), we need to find the maximum height attained and the range of the body. Given that \( g = 10 \, \text{m/s}^2 \), 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:
- \( u_x = 3 \, \text{m/s} \) (horizontal component)
- \( u_y = 10 \, \text{m/s} \) (vertical component)

### Step 2: Calculate the maximum height
The formula for the maximum height \( H \) attained by a projectile is given by:
\[ H = \frac{u_y^2}{2g} \]
Substituting the 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:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
To find \( \sin 2\theta \), we need to express \( u \) and \( \theta \) in terms of \( u_x \) and \( u_y \):
- The magnitude of the initial velocity \( u \) is:
\[ u = \sqrt{u_x^2 + u_y^2} = \sqrt{3^2 + 10^2} = \sqrt{9 + 100} = \sqrt{109} \, \text{m/s} \]
- The angle \( \theta \) can be found using:
\[ \tan \theta = \frac{u_y}{u_x} = \frac{10}{3} \]
Thus, \( \sin 2\theta = 2 \sin \theta \cos \theta \).

Using the relationships:
\[ \sin \theta = \frac{u_y}{u} = \frac{10}{\sqrt{109}}, \quad \cos \theta = \frac{u_x}{u} = \frac{3}{\sqrt{109}} \]
We can find \( \sin 2\theta \):
\[ \sin 2\theta = 2 \cdot \frac{10}{\sqrt{109}} \cdot \frac{3}{\sqrt{109}} = \frac{60}{109} \]

Now substituting \( \sin 2\theta \) into the range formula:
\[ R = \frac{u^2 \cdot \frac{60}{109}}{g} = \frac{109 \cdot \frac{60}{109}}{10} = \frac{60}{10} = 6 \, \text{m} \]

### Final Results
- Maximum Height \( H = 5 \, \text{m} \)
- Range \( R = 6 \, \text{m} \)

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

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