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JEE Main 2019 · 10 April · Shift I

JEE Main 2019 Physics question on Kinematics

A plane is inclined at an angle α=30°\alpha = 30° with respect to the horizontal. A particle is projected with a speed u=2 ms−1u = 2 \text{ ms}^{-1}, from the base of the plane, making an angle θ=15°\theta = 15° with respect to the plane as shown in the figure. The distance from the base, at which the particle hits the plane is close to

[Take, g=10 ms−2g = 10 \text{ ms}^{-2}]

  1. A.
    26 cm
  2. B.
    20 cm
  3. C.
    18 cm
  4. D.
    14 cm
Show answer and solution

Correct option: B

Solution

Using the inclined plane projectile formula, range along the plane is:

R = 2u²sin θ cos(θ + α) / g cos²α

Here θ = 15° (angle with plane), α = 30°, so (θ + α) = 45°.

R = 2(4) × sin15° × cos45° / (10 × cos²30°)

sin15° ≈ 0.2588, cos45° = 0.7071, cos²30° = 0.75

R = 8 × 0.2588 × 0.7071 / (10 × 0.75) = 8 × 0.1830 / 7.5 = 1.464 / 7.5 ≈ 0.195 m ≈ 20 cm

[JEE Main 2019 · 10 April · Shift I]

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