Posted by **Becky** on Sunday, October 11, 2009 at 11:37pm.

A test rocket is fired vertically upward from a well. A catapult gives it an initial speed of 80.2 m/s at ground level. Its engines then fire and it accelerates upward at 4.20 m/s2 until it reaches an altitude of 960 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of -9.80 m/s2. (You will need to consider the motion while the engine is operating separate from the free-fall motion.)

(a) How long is the rocket in motion above the ground?

(b) What is its maximum altitude?

(c) What is its velocity just before it collides with the Earth?

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For A, I solved it such that it was the sqrt of the initial velocity squarted + 2(acceleration times final height of rocket) = 43.8 S, but that was incorrect. For B and C, I'm a bit lost as to what to do.

- Physics -
**MathMate**, Monday, October 12, 2009 at 10:49am
If you use

S=v0*t + (1/2)at²

you have

S=960m

v0=80.2 m/s

a=4.2 m/s²

Solve for t.

I get 9.6 sec. for going up, and 14 sec. for coming down (v0=0,a=-g,S=-960).

Check my work and post if you would like to check answers.

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