Question

A projectile of mass M is fired so that the horizontal range is 4 km. At the highest poin the projectile explodes in two parts of masses M/4 and 3M/4 respectively and the heavier part start falling down vertically with zero initial speed. The horizontal range of the lighter mass will be?

Answer 1

Mass of projectile M
Initial speed = u        angle of projection = Ф
R = Range = 4,000m = u² sin2Ф / g
       =>  u² sin2Ф = 4,000 g      --- (1)

Maximum height of the trajectory = H
     H = R/4 * tanФ = 1000 tanФ    ---(2)
Horizontal distance already covered by the projectile = R/2 = 2000m

At the highest point of trajectory, vertical velocity = 0.  and the Horizontal speed = u cosФ.   Apply the conservation of linear momentum during the explosion.
 
    As the momentum in the vertical direction before explosion is 0.  It is 0 after explosion.  The heavier piece has no vertical momentum just after collision.  Hence, the lighter part also has no vertical momentum just after collision.
 
   The heavier part has no horizontal momentum after collision as it falls vertically down.   Let the lighter part have a horizontal velocity v.

     M* u cosФ = 3M/4 * 0 + M/4 * v
     v = 4 u cosФ    --- (3)

The time duration t before the lighter mass reaches ground:
       H = 0 t + 1/2 g t^2
       t^2 = 2 H/g = (2,000/g) TanФ
       t = 20 √(5 tanФ /g)  sec    --- (4)

Horizontal Distance traveled by lighter mass after explosion = v t
       vt  = 80 u cosФ √5tanФ/g)
            =  80 u √(5sinФ cosФ/g) 
            = 40 √(10 u² sin2Ф/g)            -- -(5)
            = 40  √[10*4000g/g]           substituting from (1)
            = 40 *200 = 8 km

Total range of the lighter part = 10 km