The horizontal range is equal to two times maximum height of a projectile. The angle of projection ofthe projectile is:

Options

A. `theta = tan^(-1) (4)`
B. `theta = tan^(-1) (2)`
C. `theta =45^(@)`
D. `theta = tan^(-1) ((1)/(4))`

Explanation

To solve the problem where the horizontal range \( R \) is equal to two times the maximum height \( H_{\text{max}} \) of a projectile, we can follow these steps:

### Step 1: Write down the formulas for range and maximum height.
The formulas for the horizontal range \( R \) and maximum height \( H_{\text{max}} \) of a projectile are given by:
\[ R = \frac{U^2 \sin 2\theta}{g} \]
\[ H_{\text{max}} = \frac{U^2 \sin^2 \theta}{2g} \]
where \( U \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

### Step 2: Set up the equation based on the given condition.
According to the problem, we have:
\[ R = 2 H_{\text{max}} \]
Substituting the formulas for \( R \) and \( H_{\text{max}} \):
\[ \frac{U^2 \sin 2\theta}{g} = 2 \left(\frac{U^2 \sin^2 \theta}{2g}\right) \]

### Step 3: Simplify the equation.
The \( g \) terms cancel out, and we can simplify the equation:
\[ \sin 2\theta = \sin^2 \theta \]

### Step 4: Use the double angle identity.
Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Substituting this into the equation gives:
\[ 2 \sin \theta \cos \theta = \sin^2 \theta \]

### Step 5: Rearrange the equation.
Rearranging the equation, we get:
\[ \sin^2 \theta - 2 \sin \theta \cos \theta = 0 \]
Factoring out \( \sin \theta \):
\[ \sin \theta (\sin \theta - 2 \cos \theta) = 0 \]

### Step 6: Solve for \( \theta \).
The solutions to this equation are:
1. \( \sin \theta = 0 \) (which gives \( \theta = 0^\circ \) or \( 180^\circ \), not valid for projectile motion)
2. \( \sin \theta - 2 \cos \theta = 0 \) which gives:
\[ \sin \theta = 2 \cos \theta \]
Dividing both sides by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)):
\[ \tan \theta = 2 \]

### Step 7: Find the angle of projection.
Taking the inverse tangent:
\[ \theta = \tan^{-1}(2) \]

### Final Answer:
The angle of projection \( \theta \) is \( \tan^{-1}(2) \).

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