AP EAMCET2015PhysicsMotion in Two Dimensions
A particle aimed at a target, projected with an angle 15^ with the horizontal is short of the target by 10 ~m . If projected with an angle of 45^ is away from the target by 15 ~m , then the angle of projection to hit the target is
Options
- A1 2 ⁻¹ ( 1 10 )
- B1 2 ⁻¹ ( 3 10 )
- C1 2 ⁻¹ ( 9 10 )
- D1 2 ⁻¹ ( 7 10 )
Correct answer
D. 1 2 ⁻¹ ( 7 10 )
Step-by-step solution
As Range 2 So, R-10 R+15 = 2 ₁ 2 ₂ = 30^ 90^ = 1 2 2 R-20=R+15 R=35 For the range to be maximum 2 = 90^ =1 So, R_ = u^2 g As R₁= u^2 2 g 50= u^2 g u^2=50 g and 35= 50 g 2 g [ u=50 g] 2 = 35 50 = 7 10 2 = ⁻¹ [ 7 10 ] = 1 2 ⁻¹ [ 7 10 ]