In a volcanic eruption, a 2400-kg boulder is thrown upward from ground level wit

Question

In a volcanic eruption, a 2400-kg boulder is thrown upward from ground level with an initial speed of 120 m/s at an angle of 37degree from the vertical. At the very top of its motion the boulder explodes into two pieces due to gases trapped in its ulterior. One piece is three tunes more massive than the other piece The heavier piece simply drops vertically straight down from rest and hits the ground (assumed level). How far from the initial launch point at ground level does the lighter piece hit the ground?

Explanation / Answer

a: as there is no external eenrgy applied, center of mass of exploded objects will have in the same path as that of before explosion

Range R = u^2 sin2 theta/g

Rnage = 120^2 * sin (37)/9.8

R = 1411.2 m

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massive object will reach the gprund at a distance of R/2

x1 = R/2 = 1411.02/2 = 570.5 m

centre of mass = 3m1 x1 + mx2/(3m1m2) = 1411,02

x2 = (4 * 1411.03 - 3* 570.51

x2 = 3932.5 m

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