rocket of mass 4.50 x 10^5 kg in flight, thrust directed at angle of 55.0 degrees above horizon...

Question

Question

Find magnitude and direction (as an angle above the horizontal) of rockets acceleration.

Answer 1

Answer #1

a = F / m

a = 7.5 * 10⁶ / 4.5 * 10⁵ = 16.67 m/s²

vertical component, sin 55 * 16.67 = 13.65 m/s²

13.65 - 9.8 = 3.85 m/s²

horizontal component, cos 55 * 16.67 = 9.56 m/s²

magnitude = sqrt [3.85² + 9.56²]

= 10.3 m/s²

direction theta = tan^-1(3.85 / 9.56)

= 21.9 deg

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Answer 2

Answer #2

your question is very interesting, Thanks for using homeworklib.

Answer 3

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