Question

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!

Answer 1

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