An object of mass 2 m is projected with a speed of 100 ~ms ⁻¹ at an angle = ⁻¹ ( 3 5 ) to the horizontal. At the highest point, the object breaks into two pieces of same mass m and the first one comes to rest. The distance between the point of projection and the point of landing of the bigger piece (in metre) is (Given, g=10 ~m / s ^2 )

Step-by-step solution

Horizontal range of the object fired, R= u^2 2 g At the highest point, when object is exploded into two equal masses, then 2 m u =m(0)+m v or v=2 u It means, the horizontal velocity becomes double at the highest point, hence it will cover double the distance during the remaining flight. Total horizontal range of the other part aligned & = R 2 +R= 3 R 2 & = 3 2 u^2 2 g = 3 2 (100)^2 2 g & = 3 2 (100)^2 2 3 5 4 5 10 =1440 ~m aligned