A rocket is fired vertically upward. When it reaches an altitude of 1570 m and a speed of 285 m/s........

A rocket is fired vertically upward. At the instant it reaches an altitude of 1570 m and a speed of 285 m/s, it explodes into three equal fragments.

One fragment continues to move upward with a speed of 201 m/s following the explosion.
The second fragment has a speed of 338 m/s and is moving east right after the explosion.

What is the magnitude of the velocity of the third fragment? Answer in units of m/s

By conservation of momentum, momentum_initial = momentum_final.

In the x direction, momentum_initial = 0 = 338(mass/3) + v_x(mass/3)
There v_x of the 3rd fragment = -338m/s (to the west)

In the y direction,
momentum_initial = 285*(mass) = 201(mass/3) + 0(mass/3) + v_y(mass/3) = momentum_final
285*m - 201*(m/3) = v_y(m/3)
divide by m and throughout,
285 - 201/3 = v_y/3
Therefore v_y = 654m/s (upwards)

magnitude of velocity of third fragment = (v_x ^2 + v_y ^2)^(1/2)
v = 736.179m/s

Using conservation of momentum.

The velocity of the third fragment is v at angle θ south of west.
In the east-west direction
Mv cos θ = M 338.
v cos θ = 338 .

In the up and down direction

Mv sin θ = 3M*285
v sin θ = 855

Hence
Tan θ = 855/ 338.
θ = 68.43°

v = 338/ cos 68.43° = 855/ sin 68.43° = 919.4 m/s
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