How to find the final velocity and acceleration?

Question

A 1000kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16s, then the motor stops. The rocket altitude 20s after launch is 5100m. You can ignore any effects of air resistance.

What was the rocket's acceleration during the first 16s?

What is the rocket's speed as it passes through a cloud 5100m above the ground?

Answer 1

During first 16 seconds,H=1/2A(T)^2

''''''''''''''''' last   4  seconds h=-1/2at^2+v16 t

then 5100=128A+64A-78.4====>A=26.97 m/sec^2

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v5100=-9.8(4)+26.97(16)=392.32 m/sec

Answer 2

The rocket's altitude and velocity 16 s after the launch:

h1 = (1/2)a t² = = 16²a/2 =128 a

v1 = at = 16 a

The second phase of the rocket's motion is a free fall, described by the equations:

h = -(1/2)g t² + v1 t + h1

v = -gt + v1

t= 0 is the instant when the rocket's motor stops providing thrust.

20 s after launch, that is when t = 20 - 16 = 4 s, h = 5100 m

5100 = -(1/2)*9.8*4² + 16a*4 +128a = -78.4 + 192 a

a = 5178.4 / 192 =  27 m/s ²

v = -9.8*4 + 16a = -9.8*4 + 16*27

v = 393 m/s

Answers:

The rocket's acceleration during the first 16 seconds is a = 27 m/s²

The rocket's velocity at 5100 m altitude (20 s after launch) is v = 393 m/s

Answer 3

answer to 1st question is 51/2m/s^2 and to 2nd is 510m/s