Answer to Question #1182 in Mechanics | Relativity for Jon Allen
A rocket is fired vertically upward.&
At theinstant it reaches an altitude of 1630 m and a
speed of 277 m/s, it explodes into three equal
One fragment continues to moveupward with a speed of 212 m/s following the
The second fragment has a speedof 475 m/s and is moving east right after the
What is the magnitude of the velocity ofthe third fragment?
Answer in units of m/s.
Let's write the Momentum conservation law for the horizontal and vertical projections: V: 3MV = MV1 + MV3v H: 0 = MV2 + MV3h where V = 277 m/s, V2 = 475 m/s Thus |V2| = |V3h| and |V3v| = |3V-V1| V3 = √ (V3h2 + V3v2) = √ ( 4752 + (3*277 - 212)2) = 780.25