Question

The maximum range of a projectile is 80 m. If the projectile is projected with the same speed at an angle of \( \frac{\pi}{12} \) with the horizontal, then the range of the projectile is:

• A. \(\mathbf{40 \text{ m}}\)
• B. \(80 \text{ m}\)
• C. \(20 \text{ m}\)
• D. \(60 \text{ m}\)

Hint:

For projectile motion: - The maximum range occurs at \( 45^\circ \) and is given by \( R_{\max} = \frac{u^2}{g} \). - To find range at any other angle \( \theta \), use \( R = R_{\max} \sin 2\theta \). - Use standard values of trigonometric functions to simplify calculations.

Answer 1

The Correct Option is A

Step 1: Understanding projectile motion The range \( R \) of a projectile launched at an angle \( \theta \) is given by: \[ R = \frac{u^2 \sin 2\theta}{g} \] The maximum range \( R_{\max} \) occurs when the projectile is launched at \( 45^\circ \) or \( \frac{\pi}{4} \): \[ R_{\max} = \frac{u^2}{g} \] Given that the maximum range is 80 m, we set: \[ \frac{u^2}{g} = 80 \] Step 2: Finding the range at \( \frac{\pi}{12} \) For an angle \( \theta = \frac{\pi}{12} \), the range is: \[ R' = \frac{u^2 \sin 2\theta}{g} \] Substituting \( 2\theta = \frac{\pi}{6} \): \[ R' = 80 \times \sin \frac{\pi}{6} \] Since: \[ \sin \frac{\pi}{6} = \frac{1}{2} \] \[ R' = 80 \times \frac{1}{2} = 40 \text{ m} \] Step 3: Verifying the correct option Comparing with given options, the correct answer is: \[ \mathbf{40 \text{ m}} \]