Question

a projectile is fired with velocity U at an angle of theta with the horizontal at the highest point of its trajectory its lit up into three segments of masses m m and 2 m first part Falls vertically downwards with zero initial velocity and second part returns at same path to the point of projection the velocity of third part of mass 2m just after explosion will be

Answer 1

Projectile at:  speed = u , angle = Ф,  mass = m+m+2m = 4m

At the highest point:  vertical speed = 0.   Horizontal speed = u CosФ.

The part retracing the path will have horizontal speed = u CosФ.

Apply the momentum conservation at the highest point.
           m * 0 + m * (- u CosФ) + 2 m * v = 4m * u CosФ
           v = 5/2 * u CosФ

Answer 2

Answer:5/2

Explanation:

refer pic ..