Question

If a stone is projected at certain angle ‘θ’ with the horizontal such that its horizontal range is equal to its maximum height, then angle of projection ‘θ’ is

• A. \(\tan^{-1}(2)\)
• B. \(\tan^{-1}(3)\)
• C. \(\tan^{-1}(4)\)
• D. \(\tan^{-1}(1)\)

Hint:

If range = max height in projectile motion, then \(\tan\theta = 4\), hence \(\theta = \tan^{-1}(4)\)

Answer 1

The Correct Option is C

We are given that: \[ \text{Range (R)} = \text{Maximum Height (H)} \] For a projectile: \[ R = \frac{u^2 \sin 2\theta}{g}, \quad H = \frac{u^2 \sin^2\theta}{2g} \] Set \(R = H\): \[ \frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin^2\theta}{2g} \Rightarrow 2 \sin 2\theta = \sin^2\theta \] Using identity: \(\sin 2\theta = 2\sin\theta\cos\theta\), \[ 2(2\sin\theta\cos\theta) = \sin^2\theta \Rightarrow 4\sin\theta\cos\theta = \sin^2\theta \Rightarrow 4\cos\theta = \sin\theta \Rightarrow \tan\theta = 4 \Rightarrow \theta = \tan^{-1}(4) \]

Answer 2

Given:
- A stone is projected at an angle \( \theta \) with the horizontal.
- Its horizontal range \( R \) is equal to its maximum height \( H \).

What is asked?
- Find the angle of projection \( \theta \).

Concepts and formulas involved:
For projectile motion (neglecting air resistance),
- Horizontal range:
\[ R = \frac{u^2 \sin 2\theta}{g} \] - Maximum height:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \] where,
\( u \) = initial velocity,
\( g \) = acceleration due to gravity.

Given condition:
\[ R = H \] Substitute the expressions for \( R \) and \( H \):
\[ \frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin^2 \theta}{2g} \] Cancel \( u^2 \) and \( g \) from both sides:
\[ \sin 2\theta = \frac{\sin^2 \theta}{2} \] Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \), so:
\[ 2 \sin \theta \cos \theta = \frac{\sin^2 \theta}{2} \] Multiply both sides by 2:
\[ 4 \sin \theta \cos \theta = \sin^2 \theta \] Rewrite as:
\[ 4 \sin \theta \cos \theta - \sin^2 \theta = 0 \] Factor \( \sin \theta \):
\[ \sin \theta (4 \cos \theta - \sin \theta) = 0 \] Since \( \sin \theta = 0 \) corresponds to no projectile motion, discard this solution.
Focus on:
\[ 4 \cos \theta - \sin \theta = 0 \] Rearranged:
\[ 4 \cos \theta = \sin \theta \] Divide both sides by \( \cos \theta \) (non-zero for projectile motion):
\[ 4 = \tan \theta \] Therefore:
\[ \theta = \tan^{-1}(4) \] Final answer:
\[ \boxed{\theta = \tan^{-1}(4)} \]