BITSAT2024PhysicsMotion in Two DimensionsActual
The range of the projectile projected at an angle of 15^ with horizontal is 50 m . If the projectile is projected with same velocity at an angle of 45^ with horizontal, then its range will be :
Options
- A50 m
- B50 2 ~m
- C100 m
- D100 2 ~m
Correct answer
C. 100 m
Step-by-step solution
Range of projectile R= v^2 2 g ( R (2 )) R₁ R₂ = (2 ₁ ) (2 ₂ ) = (2 15) (2 45) = 30^ 90^ 50 R ₂ = 1 2 R ₂=100 ~m