JEE Main201912 Apr 2019Morning ShiftPhysicsMotion in Two DimensionsActual
The trajectory of a projectile near the surface of the earth is given as y = 2 x - 9 x 2 . If it were launched at an angle θ 0 with speed v 0 then g = 10 m s - 2 :
Options
- Aθ 0 = cos - 1 1 5 and v 0 = 5 3 m s - 1
- Bθ 0 = cos - 1 2 5 and v 0 = 3 5 m s - 1
- Cθ 0 = sin - 1 1 5 and v 0 = 5 3 m s - 1
- Dθ 0 = sin - 1 2 5 and v 0 = 3 5 m s - 1
Correct answer
A. θ 0 = cos - 1 1 5 and v 0 = 5 3 m s - 1
Step-by-step solution
y = 2 x - 9 x 2 Comparing above equation with general equation of trajectory of projectile y = x tan ⁡ θ 1 - x R t a n   θ = 2 ⇒ s i n   θ = 2 5 or c o s   θ = 1 5 θ = s i n - 1 2 5 or θ = c o s - 1 1 5 And R = 2/9 R = v 0 2 sin ⁡ 2 θ g = 2 9 After substituting the value for g and sin 2θ. We get v 0 = 3 5 m/s