10. A rocket is launched vertically upward in the +y direction. For 27 s the roc

Question

10. A rocket is launched vertically upward in the +y direction. For 27 s the rocket's engines provide acceleration of +31.2 m/s2. What will happen the instant the rocket engines shut off? a) The rocket enters free fall, with acceleration of a-9.8 m/s. It will immediately stop and start travelling in the -y direction and fall until it hits the ground. The rocket will still travel in the +y direction for an unknown time but is not considered to be in free fall until its direction changes to be in the y direction The rocket is in free fall, with an acceleration of +9.8 m/s because it is travelling upward. It will travel in the +y direction until it reaches its maximum height. Once it stops at its maximum height the acceleration will become -9.8 m/s2 and it will travel in the -y direction. b) d) The rocket enters free fall, with acceleration of a -9.8 m/s.It will travel in the ty direction until its velocity is zero at its maximum height. Then it will travel in the -y direction until it hits the ground. Not enough information is given to answer this problem. 9 e)

Explanation / Answer

As engine shuts off, rocket will be in free fall.
free fall means acceleration will always be 9.8 m/s^2 downwards.

so a = -9.8 m/s^2 (always)

it has upward velocity when engine shuts off so it will travel upward until its speed becomes zero then it will fall downward untill crashed into the ground.

Ans(d)

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