Question

A rocket is fired vertically upward. At the
instant it reaches an altitude of 2300 m and a
speed of 279 m/s, it explodes into three equal
fragments. One fragment continues to move
upward with a speed of 224 m/s following the
explosion. The second fragment has a speed
of 358 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.

Answer 1

initial momentum up = (m1 + m2 +m3) (279) = Pi
initial momentum horizontal = 0
Final momentum up = m1 *224 + m3 v3up = Pi
Final momentum east = m2 *358 + m3 * v3e
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now all those masses are the same. I am going to call them each 1 kg
3 (279) = 1*224 + 1 * v3up
well, v3 up = 3 *279 - 224
and
v3 east = -358