Horizontal range =R=gu2sin2θ When angle of projection is (45∘−θ) then R1=gu2sin2(45∘−θ)=gu2sin(90∘−θ)=gu2cos2θ Now R2R1=u2cosg2θu2cos82θ=1:1
For angles of projection of projectile at angle (45°-θ ) and (45°+θ ), the horizontal range described by the projectile are in the ratio of:
NEET UG 2006 — Physics Mechanics
For angles of projection of projectile at angle (45∘−θ) and (45∘+θ), the horizontal range described by the projectile are in the ratio of:
Held on 30 Apr 2006 · Verified 9 Jul 2026.
Options
- A
2:1
- B
1:1
- C
2:3
- D
1:2
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