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 solve the problem of finding the maximum height attained by a body projected with a velocity \( \vec{v} = (3 \hat{i} + 4 \hat{j}) \, \text{m/s} \) and given \( g = 10 \, \text{m/s}^2 \), we can follow these steps:

### Step 1: Identify the components of the initial velocity
The velocity vector is given as:
\[ \vec{v} = 3 \hat{i} + 4 \hat{j} \, \text{m/s} \]
Here, the \( x \)-component of the velocity \( v_x = 3 \, \text{m/s} \) and the \( y \)-component of the velocity \( v_y = 4 \, \text{m/s} \).

### Step 2: Use the formula for maximum height
The formula for the maximum height \( h \) attained by a projectile is given by:
\[ h = \frac{u_y^2}{2g} \]
where \( u_y \) is the initial vertical component of the velocity and \( g \) is the acceleration due to gravity.

### Step 3: Substitute the values into the formula
In our case, \( u_y = 4 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \). Substituting these values into the formula gives:
\[ h = \frac{(4)^2}{2 \times 10} \]

### Step 4: Calculate the maximum height
Calculating the above expression:
\[ h = \frac{16}{20} = 0.8 \, \text{m} \]

### Conclusion
Thus, the maximum height attained by the body is:
\[ \boxed{0.8 \, \text{m}} \]