Posted by **blair** on Thursday, October 8, 2009 at 10:51pm.

A test rocket is fired vertically upward from a well. A catapult gives it an initial speed of 80.8 m/s at ground level. Its engines then fire and it accelerates upward at 3.90 m/s2 until it reaches an altitude of 920 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?

__________s

(b) What is its maximum altitude?

_______km

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

______m/s

Please explain this question and show what equations should be used ! Thanks!

- physics -
**MathMate**, Thursday, October 8, 2009 at 11:52pm
You have three motions to consider.

Phase 1: powered by engine

With initial velocity of v0=80.8 m/s

it slows down due to gravity until a height of S=920 m. and an acceleration of a=3.9 m/s².

THe governing equation is

S = v0*t + (1/2)a*t²

Solve for t.

The velocity at the end of this phase is

v1=v0+a*t

Phase 2: deceleration until maximum height is reached, at which time velocity = 0.

Time to reach maximum height

t1 = v1/g

Additional height travelled:

S1=v1*t1 +(1/2)(-g)t1²

Phase 3: Free fall from maximum height S+S1

Use the usual equations to calculate the velocity and time

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