A rocket is fired vertically upward. At the
instant it reaches an altitude of 1110 m and a
speed of 288 m/s, it explodes into three equal
fragments. One fragment continues to move
upward with a speed of 279 m/s following the
explosion. The second fragment has a speed of 367 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.

767.6 m/s

Explanation:

We consider momentum in both x and y components

Since momentum, p=mv where m denote mass while v is velocity then

In the y direction

3m*288=279m+my

y=585 m/s

In the x direction and taking right as positive then

3m*288=367m+mx

x=497 m/s

The resultant gives the velocity of the third fragment

molecules are free to separate from each other and will take on the shape of the container. molecules move completely at random and high speeds. they vibrate in a fixed position and do not move around much. molecules expand to the edge of their container.