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: Use the time of flight and horizontal speed:
$R = u\cos\theta\cdot T$ with $T = \frac{2u\sin\theta}{g}$. Maximum height is $H = \frac{(u\sin\theta)^2}{2g}$.

Step 2: Form the ratio:
$\frac{R}{H} = \frac{2u^2\sin\theta\cos\theta/g}{u^2\sin^2\theta/(2g)} = \frac{4\cos\theta}{\sin\theta}$.

Step 3: Set equal to n:
$\frac{4}{\tan\theta} = n$ gives $\tan\theta = \frac4n$.

Step 4: Test:
At $\theta = 45^\circ$: $R/H = 4$, matching $n = 4$ and $\tan^{-1}(1) = 45^\circ$.

Final Answer:
Option (C). \[ \boxed{\tan^{-1}\left(\frac{4}{n}\right) \text{ (C)}} \]