The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x2. If it were launched at an angle θ 0 with speed v0 then (g = 10 ms–2): (a) θ₀ = cos ⁻¹(1/√(5)) and v₀ = 5/3 ms-1 (b) θ₀ = cos ⁻¹(2/√(5)) and v₀ = 3/5 ms-1 (c) θ₀ = sin ⁻¹(2/√(5)) and v₀ = 3/5 ms-1 (d) θ₀ = sin ⁻¹(1/√(5)) and v₀ = 5/3 ms-1

Answer: (a). tan θ₀ = 2 ⇒ cos θ₀ = (1)/(√5) (g)/(2v₀²cos ²θ₀) = 9 ⇒ (10)/(2v₀²/5) = 9 ⇒ v₀² = 25/9 v₀ = 53 m/s So the answer is (a).

The trajectory of a projectile near the surface of the earth is given… | JEE Main 2019 (12 Apr, Shift 1)

Q5JEE Main 2019 · 12 APR · SHIFT 1Medium

Projectile Basics & Trajectory

The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x2. If it were launched at an angle $\theta$ 0 with speed v0 then (g = 10 ms–2):

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Answer: (a)

$\tan\theta_0 = 2 \Rightarrow \cos\theta_0 = \frac{1}{\sqrt5}$
$\frac{g}{2v_0^2\cos^2\theta_0} = 9 \Rightarrow \frac{10}{2v_0^2/5} = 9 \Rightarrow v_0^2 = \frac{25}{9}$
$v_0 = \frac53$ m/s
So the answer is (a).

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Projectile Basics & Trajectory question from JEE Main 2019 (12 Apr, Shift 1)

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