A body is projected with a velocity `vecv =(3hati +4hatj) ms^(-1)` The maximum height attained by the body is: `(g=10 ms^(-2))`

Options

A. 0.8 m
B. 8 m
C. 80 m
D. 800 m

Explanation

To find the maximum height attained by a body projected with a velocity \( \vec{v} = (3 \hat{i} + 4 \hat{j}) \, \text{m/s} \), we can follow these steps:

### Step 1: Identify the vertical component of the initial velocity
The velocity vector is given as \( \vec{v} = 3 \hat{i} + 4 \hat{j} \). Here, the vertical component of the velocity (\( u_y \)) is the coefficient of \( \hat{j} \), which is \( 4 \, \text{m/s} \).

### Step 2: Use the formula for maximum height
The formula for the maximum height (\( H_{\text{max}} \)) attained by a projectile is given by:
\[ H_{\text{max}} = \frac{u_y^2}{2g} \]
where \( g \) is the acceleration due to gravity.

### Step 3: Substitute the values into the formula
We know:
- \( u_y = 4 \, \text{m/s} \)
- \( g = 10 \, \text{m/s}^2 \)

Substituting these values into the formula:
\[ H_{\text{max}} = \frac{(4)^2}{2 \times 10} = \frac{16}{20} \]

### Step 4: Simplify the expression
Now, simplifying \( \frac{16}{20} \):
\[ H_{\text{max}} = \frac{4}{5} = 0.8 \, \text{m} \]

### Conclusion
The maximum height attained by the body is \( 0.8 \, \text{m} \).

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