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Question:

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 = 10ms⁻²):

θ₀=cos⁻¹(1√(5)) and v₀=53ms⁻¹

θ₀=sin⁻¹(1/5) and v₀=53ms⁻¹

θ₀=sin⁻¹(2/5) and v₀=53ms⁻¹

θ₀=cos⁻¹(2/5) and v₀=53ms⁻¹

Solution:

Correct option is A. θ₀=cos⁻¹(1√(5)) and v₀=53ms⁻¹
Equation of trajectory is given as
y=2x−9x².. (1)
Comparing with equation:
y=xtanθ−g/(2u²cos²θ).x².. (2)
We get; tanθ=2 ∴cosθ=1/√5
Also, g/(2u²cos²θ)=9 ⇒10/(2 × 9 × (1/√5)²)=u²
u²=25/9 ⇒u=53m/s