A kinematics question?

The velocity at the end of 60 s is 20* 60 = 1200 m/s

Height it has gone is average velocity * time = 600* 60 = 36000 m

The velocity 1200 m/s reduces to zero in a time t = 1200 /9.8 = 122.45 s

Distance traveled up in this time = average velocity* time = 600 *122.45 = 73470 m

Or using s = v^2 /2g , distance = 1200^2 /19.6 =73470 m

Total vertical height = 36000 + 73470 = 109470 m

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From the top most point the time to reach the ground is t = √ (2h/g)

√ (2*109470/9.8 ) = 149.5 s

Total time of journey = 60 +122.45 + 149.5 = 332 s

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a)Divided it into three stage, From A to B rocket will be accelerated by 20 m/s2,

then it will decelerated by gravity until its highest point C. Form C its will free fall.

hAB = 1/2 a tAB^2 = 1/2 * 20 * 60^2 = 36 000 m

It's speed at B vB = a tAB = 20 * 60 = 1200 m/s

tBC = vB/g g = 10 m/s2

tBC = 1200/10 = 120 s

hBC = vB * tBC - 1/2 g tBC^2

hBC = 1200 * 120 - 1/2 * 10 * 120^2 = 72 000 m

Maximum altitude H = hAC = hAB + hBC = 36 000 + 72 000 = 98 000 m

b) From C to ground

H = 1/2 g tCD^2

98 000 = 1/2 * 10 * tCD^2

tCD^2 = 19 600

tCD = 140 s

so Total time = tAB + tBC + tCD = 60 + 120 + 140 = 320s = 5 min 20 s