A body is projected from ground with velocity u making an angle `theta` with horizontal. Horizontal range of body is four time to maximum height attained by body. Find out
(i) angle of projection
(ii) Range and maximum height in terms of u
(iii) ratio of kinetic energy to potential energy of body at maximum height

Answer

To solve the problem step by step, we will break it down into three parts as per the requirements of the question.

### Step 1: Find the angle of projection (θ)

1. **Formulas for Range and Maximum Height:**
- The horizontal range (R) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- The maximum height (H) attained by the projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

2. **Given Condition:**
- According to the problem, the range is four times the maximum height:
\[ R = 4H \]
- Substituting the formulas for R and H:
\[ \frac{u^2 \sin 2\theta}{g} = 4 \left(\frac{u^2 \sin^2 \theta}{2g}\right) \]

3. **Simplifying the Equation:**
- Cancel \(u^2\) and \(g\) from both sides:
\[ \sin 2\theta = 4 \left(\frac{\sin^2 \theta}{2}\right) \]
- This simplifies to:
\[ \sin 2\theta = 2 \sin^2 \theta \]

4. **Using the Identity:**
- Recall that \(\sin 2\theta = 2 \sin \theta \cos \theta\):
\[ 2 \sin \theta \cos \theta = 2 \sin^2 \theta \]
- Dividing both sides by 2 (assuming \(\sin \theta \neq 0\)):
\[ \sin \theta \cos \theta = \sin^2 \theta \]

5. **Rearranging:**
- This can be rearranged to:
\[ \cos \theta = \sin \theta \]
- This equality holds true when:
\[ \theta = 45^\circ \]

### Step 2: Find Range (R) and Maximum Height (H) in terms of u

1. **Substituting θ into the Formulas:**
- For the range:
\[ R = \frac{u^2 \sin 2(45^\circ)}{g} = \frac{u^2 \cdot 1}{g} = \frac{u^2}{g} \]

2. **For the maximum height:**
- Substitute θ into the height formula:
\[ H = \frac{u^2 \sin^2(45^\circ)}{2g} = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

### Step 3: Find the ratio of Kinetic Energy (KE) to Potential Energy (PE) at Maximum Height

1. **Kinetic Energy at Maximum Height:**
- At maximum height, the vertical component of velocity is zero, and only the horizontal component remains:
\[ v_x = u \cos(45^\circ) = \frac{u}{\sqrt{2}} \]
- Therefore, the kinetic energy (KE) at maximum height is:
\[ KE = \frac{1}{2} m v_x^2 = \frac{1}{2} m \left(\frac{u}{\sqrt{2}}\right)^2 = \frac{1}{2} m \cdot \frac{u^2}{2} = \frac{mu^2}{4} \]

2. **Potential Energy at Maximum Height:**
- The potential energy (PE) at maximum height is:
\[ PE = mgh = mg \cdot \frac{u^2}{4g} = \frac{mu^2}{4} \]

3. **Ratio of KE to PE:**
- The ratio of kinetic energy to potential energy is:
\[ \frac{KE}{PE} = \frac{\frac{mu^2}{4}}{\frac{mu^2}{4}} = 1 \]

### Final Answers:
(i) Angle of projection (θ) = 45°
(ii) Range (R) = \(\frac{u^2}{g}\), Maximum Height (H) = \(\frac{u^2}{4g}\)
(iii) Ratio of KE to PE at maximum height = 1

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