another physics problem?

Question

a rocket is fired vertically from a well. A catapult gives it an initial speed of 80m/s at ground level. its engine then fires and it accelerates upward at 4 m/s^2 until it reaches an altitude of 1000 m. at that point its engines fail and the rocket goes into freefall with an acceleration of -9.8 m/s^2.

How long is the rocket in motion above the ground

What is its max altitude

what is its velocity just before it collides with the earth

(You will need to consider the motion while the engine is operating separately from the free fall)

Please help

Answer 1

Its velocity at the altitude of 1000 m is given by
V^2 = U^2 + 2as
V^2 = 80^2 + 2 x 4 x 1000
V^2 = 14400
V = 120 m/s.
Time t = [v - u] / a = [120 - 80] / 4 = 10 s
-------------------------------------------------
Velocity when it reaches the ground
W^2 = V^2 - 2 g s
W^2 = 14400 - 2 g (- 1000)
W^2 = 14400 - 2 g (- 1000)
W = - 184.4 m/s [Negative since it falls down]
Time t = [v - u] / a = [- 184.4 - 120] / {-9.8} = 31.06 s.
----------------------------------------------------------
Total time = 41.06s
Its velocity just before it collides with the earth is - 184.4 m/s
============================
The maximum altitude H = W^2 / 2g
=1734.8 m
=============================

Answer 2

The altitude of the rocket is h1 = Vo*t1 + 0.5*a*t1^2; solve for t1 for h1 = 1000m and a = 4 m/sec^2 and Vo = 80 m/s. The velocity at that point will be Vf = a*t1. The rocket will continue to climb until the vertical velocity is zero; the vertical velocity is Vf - g*t2 then g*t2 = Vf, so from Vf compute the second time interval, t2. With t2 compute the additional altitude h2 = Vf*t2 - 0.5*g*t2^2. The rocket now falls a total distance of h1 + h2. The time it takes is given by h1 + h2 = 0.5*g*t3^2; solve for t3. Now add t1+t2+t3 for the total flight time, The max altitude is h1 + h2. The fall final velocity is g*t3.

Answer 3

V^2-Vo^2=2*a*(x2-x1)
<=>V^2=2*4*1000+80^2=8080<=>v=sqrt(8000+80^2) m/s
Then the velocity of the rocket at the height of 1000 is V=sqrt(8000+80^2) m/s

The max altitude is h = V^2/(2*9.8)=(8000+80^2)/(2*9.8)+1000

The altitude just before it collides with the earth:
V2=sqrt(2*9.8*h)

(calculate yourself)

Answer 4

divide the motion into three parts
going up(with engine)
going up(without engine)
going down(freefall)

ur ans are
time = 10 + 12.24 +18.81 seconds
max altitude = 1000 + 734.69 metres
velocity = 184.39 m/s