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JEE Main 201910 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Evening Shift
Subject
Physics
Chapter
Motion in Two Dimensions

JEE Main 2019 · 10 April · Evening Shift · Physics · Motion in Two Dimensions

A plane is inclined at an angle α = 30 ° with respect to the...

A plane is inclined at an angle α=30°\alpha = 30 ° with respect to the horizontal. A particle is projected with a speed u=2ms−1u = 2m{s}^{- 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=10ms−2g = 10m{s}^{- 2} )
A plane is inclined at an angle α = 30 ° with respect to the horizontal. A particle is projected with a speed u = 2m s

Options

  1. A20cm20cm
  2. B18cm18cm
  3. C14cm14cm
  4. D26cm26cm

Correct answer

A. 20cm20cm

Step-by-step solution

Range =2u2sinαcos(α+θ)gcos⁡2θ= \dfrac{2{u}^{2}sin\alpha cos \left(\alpha + \theta\right)}{g\cos^{2} \theta}
α=angleofinclination=30°\alpha = angleofinclination = 30 °
θ=angleofprojectionfrominclination=15°\theta = angleofprojectionfrominclination = 15 °
u=2m/su = 2m / s
R=2(2)2sin⁡15°cos⁡(15°+30°)gcos⁡230°=8sin⁡15°cos⁡45°gcos230°[sin⁡15°=3−122]R = \dfrac{2{\left(2\right)}^{2}\sin 15 ° \cos \left(15 ° + 30 °\right)}{g\cos^{2} 30 °} = \dfrac{8\sin 15 °\cos 45 °}{gco{s}^{2}30 °} \left[\sin 15 ° = \dfrac{\sqrt{3} - 1}{2\sqrt{2}}\right]
R=45(3−13)≈0.2mR = \dfrac{4}{5}\left(\dfrac{\sqrt{3} - 1}{3}\right) \approx 0.2m
R≈20cmR \approx 20cm
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