A particle is projected under gravity at an angle of projection `45^(@)` with horizontal. Its horizontal range is `36m`. Find maximum Height attained by particle.

Explanation

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

### Step 1: Understand the Given Information
- The angle of projection, θ = 45 degrees.
- The horizontal range, R = 36 m.
- We need to find the maximum height attained by the particle.

### Step 2: Use the Range Formula
The formula for the horizontal range (R) of a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where:
- \( u \) = initial velocity,
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( \theta \) = angle of projection.

Since \( \theta = 45^\circ \), we know that:
\[ \sin(2\theta) = \sin(90^\circ) = 1 \]

Substituting the known values into the range formula:
\[ 36 = \frac{u^2 \cdot 1}{g} \]
This simplifies to:
\[ u^2 = 36g \]

### Step 3: Use the Maximum Height Formula
The formula for the maximum height (H) attained by a projectile is given by:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]

Since \( \theta = 45^\circ \), we have:
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \]
Thus,
\[ \sin^2(45^\circ) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]

Substituting \( u^2 \) and \( \sin^2(45^\circ) \) into the height formula:
\[ H = \frac{36g \cdot \frac{1}{2}}{2g} \]

### Step 4: Simplify the Expression
Now, simplifying the equation:
\[ H = \frac{36g \cdot \frac{1}{2}}{2g} = \frac{36 \cdot \frac{1}{2}}{2} = \frac{36}{4} = 9 \, \text{m} \]

### Final Answer
The maximum height attained by the particle is **9 meters**.

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