posted by blair on .
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?
(b) What is its maximum altitude?
(c) What is its velocity just before it collides with the Earth?
Please explain this question and show what equations should be used ! Thanks!
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
Phase 2: deceleration until maximum height is reached, at which time velocity = 0.
Time to reach maximum height
t1 = v1/g
Additional height travelled:
Phase 3: Free fall from maximum height S+S1
Use the usual equations to calculate the velocity and time