Question

A stone projected with a velocity u at an angle θ with the horizontal reaches maximum height H₁. When it is projected with velocity u at an angle ((π)/(2)-θ) with the horizontal, it reaches maximum height H₂. The relation between the horizontal range R of the projectile, heights H₁ and H₂ is:

• A. \(R = 4\sqrt{H_1H_2}\)
• B. \(R = 4(H_1-H_2)\)
• C. \(R = 4(H_1+H_2)\)
• D. R = (H₁²)/(H₂²)

Hint:

For complementary angles θ and (π)/(2)-θ: sinθ ↔ cosθ Use symmetry to simplify projectile problems.

Answer 1

The Correct Option is A

Step 1: Maximum height of a projectile: H = (u²sin²θ)/(2g)
Step 2: For angle θ: H₁ = (u²sin²θ)/(2g)
Step 3: For angle ((π)/(2)-θ): H₂ = (u²cos²θ)/(2g)
Step 4: Product of heights: H₁H₂ = (u⁴sin²θcos²θ)/(4g²)
Step 5: Range of projectile: R = (u²\sin2θ)/(g) = (2u²sinθcosθ)/(g)
Step 6: R = 4√(H₁H₂)