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, we will use the equations of motion under uniform acceleration. Here’s the step-by-step solution:

### Step 1: Understand the problem
We have a particle thrown vertically upwards, and we know its velocity at one-third of its maximum height. We need to find the maximum height attained by the particle.

### Step 2: Define variables
Let:
- \( H_{max} \) = maximum height attained by the particle
- \( v \) = velocity at one-third of the maximum height = \( 10\sqrt{2} \, \text{m/s} \)
- \( g \) = acceleration due to gravity = \( 9.81 \, \text{m/s}^2 \) (approximately \( 10 \, \text{m/s}^2 \) for simplicity)

### Step 3: Use the kinematic equation
We can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:
\[ v^2 = u^2 - 2g h \]
where:
- \( v \) = final velocity at height \( \frac{H_{max}}{3} \)
- \( u \) = initial velocity (which we will express in terms of \( H_{max} \))
- \( h \) = height = \( \frac{H_{max}}{3} \)

### Step 4: Substitute known values
At height \( \frac{H_{max}}{3} \):
\[ (10\sqrt{2})^2 = u^2 - 2g \left(\frac{H_{max}}{3}\right) \]
Calculating \( (10\sqrt{2})^2 \):
\[ 200 = u^2 - \frac{2g H_{max}}{3} \]

### Step 5: Express initial velocity \( u \)
From the equation for maximum height, we know:
\[ H_{max} = \frac{u^2}{2g} \]
Thus, we can express \( u^2 \) in terms of \( H_{max} \):
\[ u^2 = 2g H_{max} \]

### Step 6: Substitute \( u^2 \) into the equation
Substituting \( u^2 \) into our earlier equation:
\[ 200 = 2g H_{max} - \frac{2g H_{max}}{3} \]
Combining the terms:
\[ 200 = 2g H_{max} \left(1 - \frac{1}{3}\right) = 2g H_{max} \left(\frac{2}{3}\right) \]
This simplifies to:
\[ 200 = \frac{4g H_{max}}{3} \]

### Step 7: Solve for \( H_{max} \)
Rearranging gives:
\[ H_{max} = \frac{200 \times 3}{4g} \]
Substituting \( g \approx 10 \, \text{m/s}^2 \):
\[ H_{max} = \frac{600}{40} = 15 \, \text{m} \]

### Final Answer
The maximum height attained by the particle is \( H_{max} = 15 \, \text{m} \).

---