A particle is thrown vertically upward. Its velocity at half of the height is 10 m/s. Then the maximum height attained by it : -
`(g=10 m//s^2)`

Options

A. 8 m
B. 20 m
C. 10 m
D. 16 m

Explanation

To solve the problem step by step, we will use the kinematic equations of motion.

### Step 1: Understand the problem
A particle is thrown vertically upward, and we know its velocity at half of the maximum height is 10 m/s. We need to find the maximum height attained by the particle.

### Step 2: Define the variables
- Let the maximum height be \( H \).
- Therefore, the height at half of the maximum height is \( \frac{H}{2} \).
- The velocity at half the height is given as \( v = 10 \, \text{m/s} \).
- The acceleration due to gravity is \( g = 10 \, \text{m/s}^2 \).

### Step 3: Apply the kinematic equation
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) is the final velocity (0 m/s at the maximum height),
- \( u \) is the initial velocity (10 m/s at half the height),
- \( a \) is the acceleration (which is -g, or -10 m/s², since gravity acts downward),
- \( s \) is the displacement (which is \( \frac{H}{2} \)).

### Step 4: Substitute the known values
Substituting the known values into the equation:
\[ 0 = (10)^2 + 2(-10)\left(\frac{H}{2}\right) \]
This simplifies to:
\[ 0 = 100 - 10H \]

### Step 5: Solve for \( H \)
Rearranging the equation gives:
\[ 10H = 100 \]
\[ H = \frac{100}{10} = 10 \, \text{m} \]

### Conclusion
The maximum height attained by the particle is \( H = 10 \, \text{m} \).