this is my first post on this forum, so hello everyone
i take it you mean that the acceleration is constant until the fuel is burnt out.
assuming acceleration is constant, we know that up to the point of fuel burn out:
initial velocity
u = 0m/s
acceleration
a = 3.2m/s
displacement
d = 1200m
start by answering part (b):
since acceleration is constant we can apply the equation:
d = ut + 1/2at^2
where t = time taken to reach 1200m. substitute the above values into give an equation in t:
1200 = 0t + 1/2(3.2)t^2
1200 = 1.6t^2
rearange to get t:
t = √1200/1.6)
t = √750
t ≈ 27.386s
now do part (a):
a is constant so we can use:
v = u + at
where v is the velocity at 1200m. substitute in above values:
v = 0 + 3.2√750
v ≈ 87.636m/s
part (c)
consider the flight of the rocket after it runs out of fuel. we know:
acceleration a ≈ -9.8m/s/s
initial velocity
u = 3.2√750m/s
velocity at maximum altitude
v= 0m/s
constant acceleration (assuming no air resistance) so use:
v^2 = u^2 + 2ad
where d is the displacement of the rocket after it runs out of fuel. substitute in values and solve for d:
0^2 = (3.2√750)^2 + 2(-9.8)d
d = (0 - (3.2√750)^2)/2(-9.8)
d = (-(3.2√750)^2)/-19.6
d = (-10.24(750))/-19.6
d ≈ 391.837m
add this displacement to the altitude of the rocket when its fuel ran out:
maximum altitude = 1200 + 391.837 = 1591.837m
part (d)
consider the flight of the rocket after it runs out of fuel. we know:
a ≈ -9.8m/s/s
u = 3.2√750m/s
v = 0m/s
use equation:
v = u + at
0 = 3.2√750 + (-9.8)t
t = (-3.2√750)/-9.8
t ≈ 8.942s
add this to the time taken to run out of fuel:
time taken to reach max altitude = 27.386 + 8.492 = 36.382s
part (e)
u = 0m/s
a ≈ 9.8m/s/s
d = 1591.837m
v = velocity at 0 altitude.
v^2 = u^2 + 2ad
v = √(u^2 + 2ad)
v = √31200.005
v ≈ 176.635m/s
total time in the air = 36.382 + 176.635/9.8 ≈ 54.406s
all of these calculations assume constant acceleration and zero air resistance.