For angles of projection of a projectile at angles 45∘−θ and 45∘+θ, the horizontal range described by the projectile are in the ratio of
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a
2 : 1
b
1 : 1
c
2 : 3
d
1 : 2
answer is B.
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Detailed Solution
Horizontal range, R=u2sin2θgFor angle of projection 45∘−θ, the horizontal range is :R1=u2sin245∘−θg=R1=u2sin90∘−2θg=u2cos2θgFor angle of projection 45∘+θ, the horizontal range is :R2=u2sin245∘+θg=u2sin90∘+2θg=u2cos2θg∴R1R2=u2cos2θ/gu2cos2θ/g=11.