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Q.

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=u2sin⁡2θgFor angle of projection 45∘−θ, the horizontal range is :R1=u2sin⁡245∘−θg=R1=u2sin⁡90∘−2θg=u2cos⁡2θgFor angle of projection 45∘+θ, the horizontal range is :R2=u2sin⁡245∘+θg=u2sin⁡90∘+2θg=u2cos⁡2θg∴R1R2=u2cos⁡2θ/gu2cos⁡2θ/g=11.
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