A projectile of mass 3m is fired from the ground with an initial speed of 50 m s ⁻¹ at an angle of 37^ with the horizontal. At its highest point, it explodes into two fragments of masses m and 2m . The lighter fragment ( m ) completely retraces its path and lands exactly at the point of projection. The distance from the point of projection where the heavier fragment ( 2m ) lands is (Take g = 10 m s ⁻² and 37^ = 3 5 )

Step-by-step solution

Let the initial velocity be u = 50 m s ⁻¹ and angle of projection be = 37^ . The horizontal and vertical components of initial velocity are: u_x = u 37^ = 50 4 5 = 40 m s ⁻¹ u_y = u 37^ = 50 3 5 = 30 m s ⁻¹ The time of flight of the projectile if it had not exploded would be: T = 2u_y g = 2 30 10 = 6 s The horizontal range of the center of mass is unaffected by the internal forces of the explosion: X_ cm = u_x T = 40 6 = 240 m Let the landing positions of the fragments of mass m₁ = m and m₂ = 2m be x₁ and x₂ respec