contestada

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.

Respuesta :

opudodennis

Answer:

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

[tex]Third=\sqrt{585^{2}+497^{2}}=767.6157893\approx 767.6 m/s[/tex]

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