Projectile is thrown in such a way that its maximum height equals to its range, the angle of projection is
Correct answer: D. None
- A. Tan-1 (45)
- B. Tan-1 (60)
- C. Tan 1 (30)
- D. None
Explanation
Maximum Height (H): H = (u²sin²θ) / 2g, where u is the initial velocity, θ is the angle of projection, and g is acceleration due to gravity.Range (R): R = (u²sin2θ) / g2. Set up the Given ConditionThe problem states that the maximum height equals the range:H = R3. Substitute the Formulas(u²sin²θ) / 2g = (u²sin2θ) / g4. SimplifyCancel out 'u²' and 'g' from both sides: sin²θ / 2 = sin2θUse the trigonometric identity sin2θ = 2sinθcosθ: sin²θ / 2 = 2sinθcosθRearrange the equation: sin²θ = 4sinθcosθDivide both sides by sinθ (assuming θ is not 0 or 180 degrees): sinθ = 4cosθDivide both sides by cosθ: tanθ = 45. Solve for θθ = tan⁻¹(4) θ ≈ 75.96 degreesAnswer:The angle of projection is approximately 75.96 degrees.
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