The trajectory of a projectile near the surface of the earth is given as y = 2x - 9x². If it were launched at an angle θ₀ with speed v₀ then (g = 10 ms⁻²)
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The trajectory of a projectile near the surface of the earth is given as y = 2x - 9x². If it were launched at an angle θ₀ with speed v₀ then (g = 10 ms⁻²)

G1 Physics · Accepted Answer

Compare y = 2x - 9x² with y = x tanθ₀ - gx²/(2v₀²cos²θ₀). Thus tanθ₀ = 2, so θ₀ = sin⁻¹(2/√5). Also cos²θ₀ = 1/5 and 9 = 10/[2v₀²(1/5)] = 25/v₀², giving v₀ = 5/3 m/s. Therefore option C. The printed source answer key marks B, but direct comparison with the printed trajectory gives C.