A body is thrown up with a speed `u`, at an angle of projection `theta` If the speed of the projectile becomes `u/sqrt2` on reaching the maximum height , then the maximum vertical height attained by the projectile is

Options

A. `(u^2)/(4g)`
B. `(u^2)/(3g)`
C. `u^2/(2g)`
D. `(u^2)/g`

Answer

To solve the problem step by step, we will analyze the motion of the projectile and use the given information to find the maximum height attained.

### Step 1: Understand the components of the initial velocity
The initial velocity `u` can be broken down into two components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

### Step 2: Analyze the condition at maximum height
At the maximum height of the projectile, the vertical component of the velocity becomes zero. The speed of the projectile at this point is given as \( \frac{u}{\sqrt{2}} \). Since the vertical velocity is zero at maximum height, this speed must be the horizontal component of the velocity.

### Step 3: Set up the equation for horizontal velocity
Since there is no horizontal acceleration (no forces acting in the horizontal direction), the horizontal component of the velocity remains constant throughout the motion:
\[ u \cos \theta = \frac{u}{\sqrt{2}} \]

### Step 4: Solve for \( \cos \theta \)
Dividing both sides by \( u \) (assuming \( u \neq 0 \)):
\[ \cos \theta = \frac{1}{\sqrt{2}} \]

### Step 5: Determine the angle \( \theta \)
The value \( \cos \theta = \frac{1}{\sqrt{2}} \) corresponds to:
\[ \theta = 45^\circ \]

### Step 6: Use the formula for maximum height
The formula for the maximum height \( H \) attained by a projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

### Step 7: Substitute \( \theta \) into the height formula
Now, substituting \( \theta = 45^\circ \) into the formula:
- \( \sin 45^\circ = \frac{1}{\sqrt{2}} \)
Thus,
\[ H = \frac{u^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} \]

### Step 8: Simplify the expression
Calculating \( \left(\frac{1}{\sqrt{2}}\right)^2 \):
\[ H = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

### Final Answer
The maximum vertical height attained by the projectile is:
\[ H = \frac{u^2}{4g} \]

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