Question

In the projectile motion of an object, the object reaches its maximum height where its speed is half of initial speed. Then the ratio between range and maximum height of projectile is

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

Hint:

At maximum height in projectile motion, the vertical velocity becomes zero and only the horizontal velocity \(u\cos\theta\) remains. Use this condition first to find the angle of projection.

Answer 1

The Correct Option is C

Step 1: Determine projection angle from speed condition.
At max height, speed = u cosθ = u/2 → cosθ = 1/2 → θ = 60°.

Step 2: Use formulas for R and H.
R = u² sin(120°)/g = (√3 u²)/(2g). H = u² sin²(60°)/(2g) = (3u²)/(8g).

Step 3: Compute ratio R/H.
R/H = [√3/(2g)] × [8g/3] = 4√3/3 = 4/√3.

Step 4: Final Answer:
4/√3.