The projectile at the top point of the trajectory at a height of h = 100m exploded into 2 parts: M1 = 1kg and M2

The projectile at the top point of the trajectory at a height of h = 100m exploded into 2 parts: M1 = 1kg and M2 = 1.5kg. the speed of the projectile at this point is V0 = 100m / s. The speed of the larger falcon V2 turned out to be horizontal, coinciding in direction with V0 and equal to 250 m / s. Determine the distance S between the points of impact of both fragments. disregard air resistance.

If the flight starts horizontally, then the fall times of the first and second fragments are the same and equal to:
t = √ (2 * Ho / g) = √ (2 * 100 / 9.81) = 4.5152 s.
The speed of the first fragment is determined from the equality of impulses:
m * a * V * a = m * a * V * a;
V₁ = (m₂ * V₂) / V₁ = 1.5 * 250/1 = 375 m / s.
But the initial speed of 100 m / s must be subtracted:
V₁¹ = 375 – 100 = 275 m / s.
Distance L = V * t
L₁ = 275 * 4.5152 = 1241.7 m
L₂ = 250 * 4.5152 = 1128.8 m
S = L₁ + L₂ = 1241.7 + 1128.8 = 2370.5 m.