Projectile Motion – Finding the Angle of Projection from Range–Height Ratio A projectile is launched with speed u at an angle θ to the horizontal on level ground. If its horizontal range R is 2.5 times its greatest height H, determine the angle of projection θ (in degrees).

Difficulty: Easy

Correct Answer: 58°

Explanation:


Introduction / Context:
Projectile motion relationships connect range, maximum height, and launch angle. Recognizing and using standard formulas allows quick back-calculation of the launch angle when given a ratio such as R/H, a common exam style in engineering mechanics and physics.


Given Data / Assumptions:

  • Horizontal range: R = (u^2 / g) * sin 2θ.
  • Greatest height: H = (u^2 * sin^2 θ) / (2 g).
  • Given R = 2.5 * H.
  • Neglect air resistance; level launch and landing elevations.


Concept / Approach:

Form the ratio R/H to eliminate u and g, leaving a relation purely in θ. Solve for θ using trigonometric identities, specifically sin 2θ = 2 sin θ cos θ, which simplifies the ratio to a function of cot θ.


Step-by-Step Solution:

R/H = [(u^2/g) * sin 2θ] / [ (u^2 sin^2 θ)/(2g) ]= 2 * sin 2θ / sin^2 θ = 2 * (2 sin θ cos θ) / sin^2 θ = 4 * cot θGiven R/H = 2.5 ⇒ 2.5 = 4 * cot θ ⇒ cot θ = 0.625θ = arctan(1 / 0.625) = arctan(1.6) ≈ 58°


Verification / Alternative check:

Compute with a calculator: arctan(1.6) ≈ 58.0°. A quick sanity check: if θ = 60°, cot θ ≈ 0.577, giving R/H ≈ 2.308; slightly low, confirming 58° is more accurate for 2.5.


Why Other Options Are Wrong:

57° and 59° are near but yield R/H values of about 2.56 and 2.44 respectively. 60° gives 2.308. 62° deviates further.


Common Pitfalls:

Using H = u^2 sin θ / (2g) by mistake (missing the square); misapplying sin 2θ identity; forgetting that R/H must be dimensionless and u cancels.


Final Answer:

58°

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