Question

A particle projected with a velocity at an angle of \(30^\circ\) to the horizontal, travels a horizontal range R. If it is projected with the same velocity at \(60^\circ\) to the horizontal, its horizontal range will be

• A. \(\frac{R}{2}\)
• B. \(\frac{R}{4}\)
• C. 2R
• D. 3R
• E. The horizontal range remains R

Hint:

Range is the same for complementary angles (\(\theta_1 + \theta_2 = 90^\circ\)) when projected with the same velocity. Since \(30^\circ + 60^\circ = 90^\circ\), the range must be identical.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The horizontal range \(R\) of a projectile depends on the initial velocity \(u\) and the angle of projection \(\theta\).

Step 2: Key Formula or Approach:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]

Step 3: Detailed Explanation:
1. For \(\theta_1 = 30^\circ\):
\[ R_1 = \frac{u^2 \sin(2 \times 30^\circ)}{g} = \frac{u^2 \sin 60^\circ}{g} = R \]
2. For \(\theta_2 = 60^\circ\):
\[ R_2 = \frac{u^2 \sin(2 \times 60^\circ)}{g} = \frac{u^2 \sin 120^\circ}{g} \]
Recall the trigonometric identity \(\sin(180^\circ - \phi) = \sin \phi\):
\[ \sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ \]
Therefore:
\[ R_2 = \frac{u^2 \sin 60^\circ}{g} = R \]

Step 4: Final Answer:
The horizontal range remains R.