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QJEEM19Phfaf6: JEE Main 2019 Physics - Class 11 JEEM19Phfaf6 Solution: A plane is inclined at an angle = 30° with respect to the…

A plane is inclined at an angle α\alpha = 30° with respect to the horizontal. A particle is projected with a speed u = 2 ms–1 , from the base of the plane, making an angle θ\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 –2) JEE Main 2019 (Online) 10th April Evening Slot Physics - Motion in a Plane Question 63 English

Multiple Choice Options

  • A 14 cm
  • B 18 cm
  • C 20 cm
  • D 26 cm

Step-by-Step Verified Solution

T=2usin⁡θgcos⁡αT = {{2u\sin \theta } \over {g\cos \alpha }}

R=ucos⁡θT−12gsin⁡αT2R = u\cos \theta T - {1 \over 2}g\sin \alpha {T^2}

=ucos⁡θ2usin⁡θgcos⁡α−gsin⁡α24u2sin⁡2θg2cos⁡2α= {{u\cos \theta 2u\sin \theta } \over {g\cos \alpha }} - {{g\sin \alpha } \over 2}{{4{u^2}{{\sin }^2}\theta } \over {{g^2}{{\cos }^2}\alpha }}

=u2sin⁡2θgcos⁡α−u2sin⁡αgcos⁡2α{1−cos⁡2θ}= {{{u^2}{{\sin }^2}\theta } \over {g\cos \alpha }} - {{{u^2}\sin \alpha } \over {g{{\cos }^2}\alpha }}\left\{ {1 - \cos 2\theta } \right\}

=4×1210×32−u2sin⁡αgcos⁡2α{1−32}= {{4 \times {1 \over 2}} \over {10 \times {{\sqrt 3 } \over 2}}} - {{{u^2}\sin \alpha } \over {g{{\cos }^2}\alpha }}\left\{ {1 - {{\sqrt 3 } \over 2}} \right\}

=4103−830{1−32}= {4 \over {10\sqrt 3 }} - {8 \over {30}}\left\{ {1 - {{\sqrt 3 } \over 2}} \right\}

=453−830=83−830= {4 \over {5\sqrt 3 }} - {8 \over {30}} = {{8\sqrt 3 - 8} \over {30}}

=8(3−1)30=20 cm= {{8\left( {\sqrt 3 - 1} \right)} \over {30}} = 20\,cm

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