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 find the maximum height attained by the particle thrown vertically upward, we can follow these steps:

### Step 1: Understand the problem
We know that the velocity of the particle at half of the maximum height is 10 m/s. We need to find the maximum height (H) attained by the particle.

### Step 2: Define the variables
Let:
- \( v \) = final velocity at the maximum height = 0 m/s (because at the maximum height, the particle stops momentarily)
- \( u \) = velocity at half the height = 10 m/s
- \( h \) = half of the maximum height
- \( H \) = maximum height = \( 2h \)
- \( g \) = acceleration due to gravity = 10 m/s²

### Step 3: Use the kinematic equation
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) = final velocity (0 m/s at maximum height)
- \( u \) = initial velocity (10 m/s at half the height)
- \( a \) = acceleration (which is -g, since it is acting downwards)
- \( s \) = distance (which is \( h \) in this case)

Substituting the known values into the equation:
\[ 0 = (10)^2 + 2(-10)(h) \]
\[ 0 = 100 - 20h \]

### Step 4: Solve for h
Rearranging the equation gives:
\[ 20h = 100 \]
\[ h = \frac{100}{20} = 5 \text{ m} \]

### Step 5: Calculate the maximum height
Since \( H = 2h \):
\[ H = 2 \times 5 = 10 \text{ m} \]

### Conclusion
The maximum height attained by the particle is **10 meters**.

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