Time it takes for the rocket to fall down

[Noah159]

Homework Statement

Rocket is launched straight into the sky. Its engine's provide a constant acceleration of 40m/s^2.
1) What maximum height will it reach, if the engine fuel burns in t1=60 seconds?
2) How fast will the rocket fall into Earth?

Relevant Equations

V^2-Vo^2=2aS
V(t)=Vo+at
S(t)=Vot+0.5at^2

The first part I can answer. However, the answer to the second one differs from the one in the answer sheet.

1) The rocket first travels gaining speed (a=40m/s^2) and then after the fuel burns loses speed due to g=9.8 m/s^2.
Thus, Hmax= h1 (traveled with a=40) + h2 (traveled with -a=g=9.8)

h1=0.5(at^2)
h2= V1^2/2g (V1= at1), thus h2= (a^2*t1^2)/g

h=h1+h2 = 365.9 km.

Now the hard part - the second.
2) the entire time from start to the rocket reaching Earth once again is t=t1 (traveled with a=40) + t2 (with speed and -a=g) + t3(time it takes to free fall from hmax). t=t1+t2+t3

t1=60 sec (given)
t2= V1/g = at1/g (244.9 sec)

t3= this is the problem. I get different t3 compared to the answer sheet. The answer sheet gives t3= √(2h/g)
However, they take 2h= t1^2*a. (and get t3= t1√(a/g)
How does that make sense? The height used should be Hmax, because it falls from the very top.?

I believe it should be t3= √(2Hmax/g), thus t3= 273.3 sec.?
Help would be appreciated.

 

[etotheipi]

It's probably a mistake. If the rocket is falling from $366\text{km}$ in the third stage, you use that value.