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A 1000 kg weather rocket is launched straight up.The rocket motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 6200 m. 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 6200 m above the ground?

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Added Wed, 09 Dec '15
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The rocket's altitude and velocity 16 s after the launch:

h1 = (1/2)a t� = = 16�a/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 t� + 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 = 6200 m

6200 = -(1/2)*9.8*4� + 16a*4 +128a = -78.4 + 192 a

a = 5178.4 / 192 = 32.7 m/s �

v = -9.8*4 + 16a = -9.8*4 + 16*32.7

v = 484 m/s

Answers:

The rocket's acceleration during the first 16 seconds is a = 32.7 m/s�

The rocket's velocity at 6200 m altitude (20 s after launch) is v = 484 m/s

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Added Wed, 09 Dec '15
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