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2-Dimensional motion problem

  1. Oct 13, 2004 #1
    OK---a rocket is launched at 53 degrees above horizontal with an initial speed of 100 m/s. It moves along its initial straight line path with an acceleraqtion of 30 m/s^2 for 3 seconds. At this time, the rocket is turned off and it proceeds to fall as a free body.

    how can i find,A) the maximum height reached. B) the total time of flight. C) its horizontal range. ???????


    another problem im having is the escalator problem which the vectors move in the same direction...

    You notice you can walk up a stationary (broken) escalator in 30 seconds-- while you can go up in 20 seconds while it is working just by standing on it. How long would it take for you to WALK up the working escalator?



    any help plz?! :bugeye: :confused: :confused:
     
  2. jcsd
  3. Oct 14, 2004 #2

    HallsofIvy

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    Have you TRIED anything?

    You are told that the rocket moves along a straight line, making angle 53 degrees with the horizontal, with acceleration 30 m/s2.
    You should know:
    v = at+ v0 and x= (1/2)at2+ v0t so you can calculate the distance the rocket has moved on that line and its speed in that direction after 3 seconds.

    Convert those to x and y coordinates (using sin(53) and cos(53)) and use them as
    (x1,y1) and (vx1,vy1), new initial position and velocity vectors with acceleration -g in the y direction and 0 in the x direction.
     
  4. Oct 15, 2004 #3

    pervect

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    Here is a very large hint. Velocities add.

    The velocity of the escalator is distance ;/ time

    d / 20 seconds

    where d is some distance.

    So:

    Write a similar expression for the velocity at which you walk up the escalator.

    Add the velocities together to get the total velocity.

    Solve the formula distance = velocity * time for the time it takes to go up at the new velocity which is the sum.

    You will find you won't be able to solve for 'd', but you don't need to.
     
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