Question

The speed of a projectile at its maximum height is \( \frac{\sqrt{3}}{2} \) times its initial speed. If the range of the projectile is \( n \) times the maximum height attained by it, then \( n \) is equal to:

• A. \( \frac{4}{3} \)
• B. \( 2\sqrt{3} \)
• C. \( 4\sqrt{3} \)
• D. \( 3/4 \)

Hint:

When dealing with projectile motion, always check the relationships between range and height, especially when given conditions like horizontal velocity at maximum height.

Answer 1

The Correct Option is C

In projectile motion, the maximum height \( H \) is given by: \[ H = \frac{v_0^2 \sin^2(\theta)}{2g} \] and the range \( R \) is given by: \[ R = \frac{v_0^2 \sin(2\theta)}{g} \] The speed at the maximum height is the horizontal component, which is \( v_0 \cos(\theta) \). According to the problem, this is \( \frac{\sqrt{3}}{2} v_0 \), so: \[ \cos(\theta) = \frac{\sqrt{3}}{2} \] This implies \( \theta = 30^\circ \). Substituting \( \theta = 30^\circ \) into the range equation, we get: \[ R = \frac{v_0^2 \sin(60^\circ)}{g} = \frac{v_0^2 \cdot \frac{\sqrt{3}}{2}}{g} \] The maximum height is: \[ H = \frac{v_0^2 \sin^2(30^\circ)}{2g} = \frac{v_0^2 \cdot \frac{1}{4}}{2g} = \frac{v_0^2}{8g} \] Given that the range is \( n \) times the maximum height: \[ R = n \cdot H \] Substituting the expressions for \( R \) and \( H \), we get: \[ \frac{v_0^2 \cdot \frac{\sqrt{3}}{2}}{g} = n \cdot \frac{v_0^2}{8g} \] Simplifying: \[ \frac{\sqrt{3}}{2} = \frac{n}{8} \] Solving for \( n \): \[ n = 4\sqrt{3} \] Thus, the value of \( n \) is \( 4\sqrt{3} \).