Question

The angle of projection for a projectile to have same horizontal range and maximum height is :

• A. \(\tan^{-1}(2)\)
• B. \(\tan^{-1}(4)\)
• C. \(\tan^{-1}\left(\frac{1}{4}\right)\)
• D. \(\tan^{-1}\left(\frac{1}{2}\right)\)

Answer 1

The Correct Option is B

To determine the projection angle for equal horizontal range and maximum height, we utilize the projectile motion equations.

The horizontal range \( R \) and maximum height \( H \) are defined as:

• \(R = \frac{v^2 \sin(2\theta)}{g}\)
• \(H = \frac{v^2 \sin^2(\theta)}{2g}\)

where:

• \( v \) denotes the initial velocity.
• \( \theta \) represents the angle of projection.
• \( g \) is the acceleration due to gravity.

The condition to satisfy is \(R = H\).

Equating the expressions yields:

\(\frac{v^2 \sin(2\theta)}{g} = \frac{v^2 \sin^2(\theta)}{2g}\)

After canceling common terms and rearranging:

\(2 \sin(2\theta) = \sin^2(\theta)\)

Applying the identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \):

\(2 \times 2\sin(\theta)\cos(\theta) = \sin^2(\theta)\)

Simplifying this equation gives:

\(4 \sin(\theta) \cos(\theta) = \sin^2(\theta)\)

Dividing both sides by \( \sin(\theta) \) (assuming \( \theta eq 0 \)):

\(4 \cos(\theta) = \sin(\theta)\)

This leads to:

\(\frac{\sin(\theta)}{\cos(\theta)} = 4\)

Therefore, \( \tan(\theta) = 4 \).

The angle \( \theta \) is found to be \( \theta = \tan^{-1}(4) \).

Consequently, the required projection angle is \( \tan^{-1}(4) \). The correct selection is:

\(\tan^{-1}(4)\)