Question

The two projectiles are projected with the same initial velocities at the 15° and 30° with respect to the horizontal. The ratio of their ranges is 1:x. The value of \(x\) is

• A. \(\sqrt{2}\)
• B. \(\sqrt{3}\)
• C. \(2\sqrt{5}\)
• D. \(\frac{1}{\sqrt{2}}\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The horizontal range of a projectile on level ground is determined by its initial velocity and angle of projection.
Step 2: Key Formula or Approach:
The range \(R\) of a projectile is given by the formula:
\(R = \frac{u^2 \sin(2\theta)}{g}\)
where \(u\) is the initial velocity, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity.
Step 3: Detailed Explanation:
For the first projectile, the angle of projection is \(\theta_1 = 15^\circ\).
Its range is:
\(R_1 = \frac{u^2 \sin(2 \times 15^\circ)}{g} = \frac{u^2 \sin(30^\circ)}{g} = \frac{u^2}{2g}\).
For the second projectile, the angle of projection is \(\theta_2 = 30^\circ\).
Its range is:
\(R_2 = \frac{u^2 \sin(2 \times 30^\circ)}{g} = \frac{u^2 \sin(60^\circ)}{g} = \frac{u^2 \sqrt{3}}{2g}\).
Now, calculate the ratio \(R_1 : R_2\):
\(\frac{R_1}{R_2} = \frac{\frac{u^2}{2g}}{\frac{u^2 \sqrt{3}}{2g}} = \frac{1}{\sqrt{3}}\).
Step 4: Final Answer:
We are given that the ratio is \(1 : x\).
Comparing \(\frac{1}{x} = \frac{1}{\sqrt{3}}\), we get \(x = \sqrt{3}\).