The height reached by the rocket depends on the amount of fuel

Initial displacement = dboost = v0t + 1/2at^2 
hboost = dboost sin theta 
xboost = dboost cos theta 
vboost = v0 + at 
vxboost = vboost cos theta 
vyboost = vboost sin theta 

To get the maximum altitude, use conservation of energy 
initial PE + vertical KE 
= mg (hboost) + 1/2 m vyboost^2 
= final PE = mg (hfinal) 

hfinal = hboost + vyboost^2 / 2g 
= hboost + ((v0 + at) * sin (theta))^2 / 2g 

b) flight time = 
boost time + time to rise ballistically from hboost to hfinal 
+ time to fall from hfinal 

The time to fall a distance h can be gotten by: 
h = 1/2 gt^2, so t = sqrt (h/2g) 

So flight time = t + sqrt ((hfinal - hboost)/2g)+ sqrt (hfinal/2g) 

Horizontal range = xboost + vxboost * time of ballistic flight 

= (v0t + 1/2at^2) cos theta 
+ (v0 + at) cos theta * (sqrt ((hfinal - hboost)/2g)+ sqrt (hfinal/2g)) 

Lots of plugging and chugging to do now. 

EDIT--the problem said acceleration was constant, so we don't have to worry about loss of mass during the boost. Realistic? No. But that's what the problem says. Once the thing is ballistic, it's a moot point anyway

True as u cant have a rocket flying at a high height if its low on fuel as if the fuel runs out mid flight they will just crash

brainliest:

Arhombus is a quadrilateral.