Question

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.

Answer 1

767.6 m/s

Step-by-step 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

`Third=\sqrt{585^(2)+497^(2)}=767.6157893\approx 767.6 m/s`