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Given, u₁ = u₂ = u, θ₁ = 60°, θ₂ = 30°. In 1st case, we know that range R₁ = (u² sin60°) / g = (u² sin120°) / g = (u² sin(90° + 30°)) / g = (u² cos30°) / √3u² / 2g. In 2nd case, when θ₂ = 30°, then R₂ = (u² sin60°) / g = (u² √3) / 2g ⇒ R₁ = R₂ [we get same value of ranges].

1. Horizontal range is same when angle of projection with the horizontal is θ and (90° − θ).
2. Rmax = v² / g = 16000 [16km = 16000m].
3. v = (16000g)¹/² = (16000 × 10)¹/² = 400 ms⁻¹.
4. Velocity of swimmer w.r.t river Vₛ = 20 m/s Velocity of river w.r.t. ground Vᵣₐ₉ = 10 m/s.

Correct answer: Horizontal range is same when angle of projection with the horizontal is θ and (90° − θ).

Solution

The horizontal range of a projectile is the same for complementary angles of projection (θ and 90° − θ) when the initial speed is the same. This is a standard result derived from the range formula.

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