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While moving horizontally at 3.5 x 103 m/s at an altitude of 2.0 x 104 m, a ballistic missile explodes and breaks into two fragments which fall freely. The smaller piece, which has 35% of the missile's total mass, falls straight down after the explosion and lands at point A, the place on the ground directly below the point of the explosion. How far from point A does the other fragment land?

Respuesta :

Answer:

3, 43,968.88 m

Explanation:

Let mass of the missile be m

velocity = 3500 m/s

smaller part will have zero horizontal velocity and larger part will have velocity  v in horizontal direction

Applying conservation of momentum

m x 3500 = .35m x 0 + .65m x v

v = 5384.61 m / s

Height of  explosion

h = 20000 m

time to fall be t

for vertical fall u = 0 , g = 9.8

20000 = 1/2 x 9.8 t²

= 63.88 s

Horizontal range

time of fall x horizontal velocity

= 63.88 x 5384.61

= 3, 43,968.88 m

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