Kinematics

JEE Main level, single correct, medium

263 of 308

Question 263 of 308

A plane is inclined at an angle α=30∘\alpha =30^{\circ} 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^{\circ} 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} )

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Answer: (A) 20cm20cm

Range =2u2sinαcos⁡⁡(α+θ)gcos⁡2⁡θ=\dfrac{2{u}^{2}sin\alpha \cos ⁡\left(\alpha +\theta \right)}{g{\cos }^{2}⁡\theta }
α=angleofinclination=30∘\alpha =angleofinclination=30^{\circ}
θ=angleofprojectionfrominclination=15∘\theta =angleofprojectionfrominclination=15^{\circ}
u=2m/su=2m/s
R=2(2)2sin15∘cos⁡⁡(15∘+30∘)gcos⁡2⁡30∘=8sin⁡⁡15∘cos45∘gcos230∘R=\dfrac{2{\left(2\right)}^{2}sin15^{\circ}\cos ⁡\left(15^{\circ}+30^{\circ}\right)}{g{\cos }^{2}⁡30^{\circ}}=\dfrac{8\sin ⁡15^{\circ}cos45^{\circ}}{gco{s}^{2}30^{\circ}} [sin15∘=3−122]\left[sin15^{\circ}=\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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