Range =gcos2θ2u2sinαcos(α+θ)
α=angleofinclination=30∘
θ=angleofprojectionfrominclination=15∘
u=2m/s
R=gcos230∘2(2)2sin15∘cos(15∘+30∘)=gcos230∘8sin15∘cos45∘ [sin15∘=223−1]
R=54(33−1)≈0.2m
R≈20cm
JEE Main 2019 — Physics Mechanics
A plane is inclined at an angle α=30∘ with respect to the horizontal. A particle is projected with a speed u=2ms−1 , from the base of the plane, making an angle θ=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−2 )

Held on 10 Apr 2019 · Verified 6 Jul 2026.
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