Question

The horizontal range of a projectile (R) is given by: ____.

• A. R = u² cos 2$\alpha$ g
• B. R = u² sin 2$\alpha$ g
• C. R = u² cos $\alpha$ g
• D. R = u² sin $\alpha$ g

Hint:

The range is maximum when $\sin 2\alpha = 1$, which happens at an angle of projection of $45^\circ$. Also, the range is the same for complementary angles (e.g., $30^\circ$ and $60^\circ$).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The horizontal range ($R$) is the total horizontal distance traveled by the projectile from its point of launch to the point where it returns to the same horizontal level.

Step 2: Key Formula or Approach:
Range is the product of the constant horizontal velocity ($u \cos \alpha$) and the total time of flight ($T = 2u \sin \alpha / g$).

Step 3: Detailed Explanation:
Calculation of Range ($R$): \[ R = (u \cos \alpha) \times \left( \frac{2u \sin \alpha}{g} \right) \] \[ R = \frac{u^2 (2 \sin \alpha \cos \alpha)}{g} \] Using the trigonometric double-angle identity ($2 \sin \alpha \cos \alpha = \sin 2\alpha$): \[ R = \frac{u^2 \sin 2\alpha}{g} \]

Step 4: Final Answer:
The horizontal range is given by $R = \frac{u^2 \sin 2\alpha}{g}$.

Answer 2

This question asks for the correct formula for the horizontal range of a projectile launched at angle \( \alpha \) with speed \( u \). Rather than re-deriving it from scratch, let's test each option against known physical behavior of projectile range.

1. \( R = \dfrac{u^2\cos 2\alpha}{g} \): Using \( \cos 2\alpha \), the range would be maximum at \( \alpha = 0^\circ \) (a purely horizontal launch) and would actually turn negative for angles above 45°, which contradicts the well-known fact that a projectile achieves its maximum range at a 45° launch angle, not at 0°.
2. \( R = \dfrac{u^2\sin 2\alpha}{g} \): The function \( \sin 2\alpha \) reaches its peak value of 1 exactly when \( 2\alpha = 90^\circ \), that is, when \( \alpha = 45^\circ \). This matches the well-established physical result that a projectile travels the farthest horizontal distance when launched at 45° to the horizontal, for a fixed launch speed.
3. \( R = \dfrac{u^2\cos\alpha}{g} \): This expression would be maximum at \( \alpha = 0^\circ \), predicting that launching perfectly horizontally gives the farthest range, which does not correctly represent range as a function of launch angle in projectile motion.
4. \( R = \dfrac{u^2\sin\alpha}{g} \): This form would keep increasing all the way up to \( \alpha = 90^\circ \) (a straight vertical launch), which is physically wrong since a projectile launched straight up travels zero horizontal distance, not the maximum.

Only the option using \( \sin 2\alpha \) correctly predicts that the range peaks at a 45° launch angle and returns to zero at both 0° and 90°, matching the true behavior of projectile range.

Therefore, the correct answer is \( R = \dfrac{u^2\sin 2\alpha}{g} \).