A 300 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 26 m/s2 for 36 s, then runs out of...?

(a)
Take g = 9.81 m/s², and take all upward quantities to be positive.

The journey can be divided into 3 parts.
Part 1 : From start to the time running out of fuel
Part 2 : From the time running out of fuel to the maximum point
Part 3 : Falling to the ground from the starting point

Consider Part 1 : From start to the time running out of fuel
Initial velocity, u = 0 m/s
Acceleration, a = 26 m/s²
Time taken, t = 36 s

s = ut + (1/2)at²
s = 0 + (1/2) × 26 × 36² m
Displacement of Part 1, s = 16848 m

v = u + at
v = 0 + 26 × 36 m/s
Final velocity of Part 1, v = 936 m/s

Consider Part 2 : From the time running out of fuel to the maximum point
Initial velocity, u = 938 m/s
Final velocity, v = 0 m/s
Acceleration, a = -9.81 m/s

v = u + at
0 = 938 + (-9.81) × t
t = 938/9.81 s
Time taken for Part 2, t = 95.6 s

v² = u² + 2as
0 = 938² + 2 × (-9.81) × s
s = 938² / (2 × 9.81) m
Displacement of Part 2, s = 44844 m

Maximum altitude
= (Distance moving in Part 1) + (Distance moving in Part 2)
= (16848 + 44844) m
= 64692 m
≈ 64700 m (to 3 sig. fig.)


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(b)
Part 3 : Falling to the ground from the starting point
Initial velocity, u = 0
Displacement, s = -64692 m
Acceleration, a = -9.81 m/s

s = ut + (1/2)at²
-64692 = 0 + (1/2) × (-9.81) × t²
t = √(2 ×64692 / 9.81) s
Time taken for Part 3 = 114.8 s

Time taken before the rocket hitting the ground
= (Time taken for Part 1) + (Time taken for Part 2) + (Time taken for Part 3)
= (36 + 95.6 + 114.8)s
= 246.4 s
≈ 246 s (to 3 sig. fig.)

t1 = 36 sec
h = a/2*t1^2 = 13*36^2 = 16,848 m
V = a*t1 = 26*36 = 936 m/sec
Δh = V^2/2g = 936^2/19.612 = 44,671 m
H = h+Δh = 16,848+44,671 = 61,519 m
t2 = V/g = 936/9.806 = 95.45 sec
t3 = √2H/g = √61,519*2/9.806 = 112.01 sec
t = t1+t2+t3 = 95.45+112.01+36.00 = 243.46 sec