Find the angle of projection of a projectile for which the horizontal range and maximum height are equal.

Options

A. `theta=tan^(-1)((1)/(4))`
B. `theta=tan^(-1)(4)`
C. `theta=tan^(-1)(2)`
D. `theta =tan45^(@)`

Answer

To find the angle of projection of a projectile for which the horizontal range and maximum height are equal, we can follow these steps:

### Step 1: Understand the formulas for maximum height and horizontal range
The maximum height \( H \) of a projectile is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

The horizontal range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]

### Step 2: Set the maximum height equal to the horizontal range
According to the problem, we need to find the angle \( \theta \) such that:
\[ H = R \]
Substituting the formulas for \( H \) and \( R \):
\[ \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin 2\theta}{g} \]

### Step 3: Simplify the equation
We can cancel \( u^2 \) from both sides (assuming \( u \neq 0 \)) and \( g \) from both sides:
\[ \frac{\sin^2 \theta}{2} = \sin 2\theta \]

### Step 4: Rewrite \( \sin 2\theta \)
Using the double angle identity, we can rewrite \( \sin 2\theta \):
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
So the equation becomes:
\[ \frac{\sin^2 \theta}{2} = 2 \sin \theta \cos \theta \]

### Step 5: Rearrange the equation
Multiplying both sides by 2 to eliminate the fraction:
\[ \sin^2 \theta = 4 \sin \theta \cos \theta \]

### Step 6: Factor the equation
Rearranging gives:
\[ \sin^2 \theta - 4 \sin \theta \cos \theta = 0 \]
Factoring out \( \sin \theta \):
\[ \sin \theta (\sin \theta - 4 \cos \theta) = 0 \]

### Step 7: Solve for \( \theta \)
This gives us two cases:
1. \( \sin \theta = 0 \) (which corresponds to \( \theta = 0^\circ \) or \( 180^\circ \), not applicable for projectile motion)
2. \( \sin \theta - 4 \cos \theta = 0 \) or \( \sin \theta = 4 \cos \theta \)

Dividing both sides by \( \cos \theta \):
\[ \tan \theta = 4 \]

### Step 8: Find the angle \( \theta \)
Thus, the angle of projection \( \theta \) is:
\[ \theta = \tan^{-1}(4) \]

### Final Answer
The angle of projection for which the horizontal range and maximum height are equal is:
\[ \theta = \tan^{-1}(4) \]