Two projectiles are thrown from the ground with the same initial speed. The first projectile is thrown at an angle of 15^ with the horizontal. If the horizontal range of the second projectile is twice that of the first projectile, the angle of projection of the second projectile with the horizontal is

Step-by-step solution

Let the initial speed of both projectiles be u . The horizontal range of a projectile is given by R = u^2 2 g . For the first projectile: R₁ = u^2 (2 15^ ) g = u^2 30^ g = u^2 2g For the second projectile, the range is twice that of the first: R₂ = 2R₁ = 2 ( u^2 2g ) = u^2 g Let be the angle of projection for the second projectile. Then, R₂ = u^2 2 g Equating the two expressions for R₂ : u^2 2 g = u^2 g 2 = 1 2 = 90^ = 45^ Answer: 45^