A body is projected from the ground into the air . The velocity of the body at a height of 5 from the ground is given by `vecv=3hati+8hatj`. Calculate the maximum height attained by the body `(g=10ms^(-2))`

Answer

To solve the problem, we need to calculate the maximum height attained by a body projected into the air, given its velocity at a height of 5 meters. The velocity vector at this height is given as \( \vec{v} = 3 \hat{i} + 8 \hat{j} \) and the acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).

### Step-by-Step Solution:

**Step 1: Identify the components of the velocity vector.**
- The velocity vector \( \vec{v} = 3 \hat{i} + 8 \hat{j} \) has:
- Horizontal component \( v_x = 3 \, \text{m/s} \)
- Vertical component \( v_y = 8 \, \text{m/s} \)

**Step 2: Use the kinematic equation to find \( u^2 \sin^2 \theta \).**
- We use the kinematic equation:
\[ v_y^2 = u_y^2 - 2gh \]
where:
- \( v_y = 8 \, \text{m/s} \) (final vertical velocity at height 5 m)
- \( u_y = u \sin \theta \) (initial vertical velocity)
- \( g = 10 \, \text{m/s}^2 \)
- \( h = 5 \, \text{m} \)

- Rearranging the equation gives:
\[ 8^2 = (u \sin \theta)^2 - 2 \cdot 10 \cdot 5 \]
\[ 64 = (u \sin \theta)^2 - 100 \]
\[ (u \sin \theta)^2 = 64 + 100 = 164 \]

**Step 3: Calculate the maximum height.**
- The maximum height \( H \) is given by:
\[ H = \frac{(u \sin \theta)^2}{2g} \]
- Substituting the values:
\[ H = \frac{164}{2 \cdot 10} = \frac{164}{20} = 8.2 \, \text{m} \]

**Final Answer:**
The maximum height attained by the body is \( 8.2 \, \text{m} \).