Question

Review Part A 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 7100 m. You can ignore any effects of What was the rocket's acceleration during the first 16 s? Express your answer with the appropriate units air resistance al x.10" | a= | Value Units Submit Previous Answers Request Answer

Answer 1

(1) Suppose acceleration of rocket for first 16 seconds = a m/s^2

So, speed of the rocket after 16 s is -

v = a*t = a*16 = 16a m/s

Distance traveled in 16 s -

s1 = (1/2)*a*t^2 = 0.5*a*16^2 = 128a meter.

Now, when the rocket engine shuts down after 16 s, gravitational acceleration 'g' will come in picture.

so -

s2 = v*t' - (1/2)*g*t^2

s2 = 16a*(20-16) - 0.5*9.8*(20-16)^2

=> s2 = 64a - 78.4 meter

Given that -

s1 + s2 = 7100 m

=> 128a + 64a - 78.4 = 7100

=> 192a = 7178.4

=> a = 7178.4 / 192 = 37.4 m/s^2 (Answer)

(2) Solution in enclosed sheet.

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