Question

For a projectile motion, the range R is 'n' times the maximum height H. So the angle of projection is

• A. \(sin^{-1}(\frac{2}{n})\)
• B. \(cos^{-1}(\frac{4}{n})\)
• C. \(tan^{-1}(\frac{4}{n})\)
• D. \(tan^{-1}(\frac{2}{n})\)

Hint:

Divide R = u^2 sin2theta / g by H = u^2 sin^2theta / 2g to get R/H = 4 / tan theta.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a projectile launched with speed \(u\) at angle \(\theta\), the range and the maximum height are fixed by the same two quantities.

Step 2: Key Formula or Approach:
\[ R = \frac{u^2\sin2\theta}{g}, \qquad H = \frac{u^2\sin^2\theta}{2g} \]

Step 3: Detailed Explanation:
Divide range by height:
\[ \frac{R}{H} = \frac{u^2\cdot2\sin\theta\cos\theta/g}{u^2\sin^2\theta/2g} = \frac{4\cos\theta}{\sin\theta} = \frac{4}{\tan\theta} \]
It is given that \(R = nH\), so
\[ \frac{4}{\tan\theta} = n \Rightarrow \tan\theta = \frac{4}{n} \Rightarrow \theta = \tan^{-1}\left(\frac4n\right) \]
Check: for \(n = 4\) the angle is \(45^\circ\), which is the angle at which \(R = 4H\) for a projectile. The options in (A) and (D) have \(2\) in place of \(4\), and option (B) uses cosine, none of which gives this check.

Final Answer:
The angle of projection is \(\tan^{-1}(4/n)\), option (C). \[ \boxed{\tan^{-1}\left(\frac{4}{n}\right) \text{ (C)}} \]