When a particle is thrown vertically upwards, its velocity at one third of its maximum height is `10sqrt2m//s`. The maximum height attained by it is

Options

A. `20sqrt2m`
B. `30m`
C. 15 m
D. 12.8 m

Answer

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

### Step 1: Understand the problem
We are given the velocity of the particle at one-third of its maximum height. We need to find the maximum height attained by the particle.

### Step 2: Define the variables
Let:
- \( h \) = maximum height attained by the particle
- \( v \) = velocity at one-third of the maximum height = \( 10\sqrt{2} \, \text{m/s} \)
- The height at one-third of the maximum height = \( \frac{h}{3} \)

### Step 3: Use the kinematic equation
We will use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) = final velocity (at maximum height, this is 0)
- \( u \) = initial velocity (at one-third of the height, which is \( 10\sqrt{2} \, \text{m/s} \))
- \( a \) = acceleration (which is \( -g = -10 \, \text{m/s}^2 \) due to gravity)
- \( s \) = displacement (which is the distance from one-third height to maximum height, \( h - \frac{h}{3} = \frac{2h}{3} \))

### Step 4: Substitute the known values into the equation
Substituting the values into the equation:
\[ 0 = (10\sqrt{2})^2 + 2(-10)\left(\frac{2h}{3}\right) \]

### Step 5: Simplify the equation
Calculating \( (10\sqrt{2})^2 \):
\[ 0 = 200 - \frac{40h}{3} \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ \frac{40h}{3} = 200 \]

### Step 7: Solve for \( h \)
Multiplying both sides by 3:
\[ 40h = 600 \]
Now, divide by 40:
\[ h = \frac{600}{40} = 15 \, \text{m} \]

### Conclusion
The maximum height attained by the particle is \( 15 \, \text{m} \).

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