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Projectile motion

  1. Jun 23, 2011 #1
    1. a rocket is launched from rest and moves in a straight line at 70degree above the horizontal with an acceleration of 46ms^(-2) . After 30seconds of powered flight , the engines shut off and the rocket follows a parabolic path back to earth. find
    1)the maximun altitude reached


    2. i use this formula, s= ut +1/2at^2
    s=? , u=0, t= 30, a= (46-9.81 )
    the correct answer is 105.16km, i didt get the answer~


    3. why cant get the answer and how should i do to get correct answer?
       help me...thank you
     
  2. jcsd
  3. Jun 23, 2011 #2

    gneill

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    Staff: Mentor

    During the powered portion of the flight you are given the rocket's angle and acceleration, and told that the trajectory is a straight line. So at the end of the powered portion (20 seconds duration) what is the speed of the rocket and its height? How about the vertical component of its speed?
     
  4. Jun 23, 2011 #3
    use which one formula?
     
  5. Jun 23, 2011 #4

    gneill

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    For speed, use a formula that includes acceleration and time and gives velocity.

    For distance, use a formula that includes acceleration and time and gives distance :smile:

    What formulas do you know?
     
  6. Jun 23, 2011 #5
    i know this formula, v= u + at
    s= ut +1/2at^2, correct?
     
  7. Jun 23, 2011 #6

    gneill

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    Yes, fine. Which one will give you the speed of the rocket, traveling in a straight line with acceleration a=46ms-2 for 30 seconds?
     
  8. Jun 23, 2011 #7
    1380m/s??
    after that?
    i use this formula s=ut+1/2 at^2 to find distance
    u= 0, t=30 a=46, correct?
     
  9. Jun 23, 2011 #8

    gneill

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    Yes, fine so far. Determine the height of the rocket at burnout from its distance traveled (it's been traveling at a 70° angle, remember). From the speed and angle, determine the vertical component of the velocity. Then treat these circumstances as a new projectile problem for the unpowered portion of the flight.
     
  10. Jun 23, 2011 #9
    s=ut +1/2 at^2
    s= 1380sin 70(30)+ 1/2 (46)(30)^2
    my answer is 59.603km
    but answer is 105.16km
     
  11. Jun 23, 2011 #10

    gneill

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    Your speed, 1380m/s is at the *end* of the 30 second burn. It is not the initial velocity of the rocket (that's zero).

    You are currently looking for the height of the rocket when its fuel runs out (after 30 seconds). What is that height?
     
  12. Jun 23, 2011 #11
    then using wat formula?
     
  13. Jun 23, 2011 #12

    gneill

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    Using the formula that gives you distance when you know the acceleration and time.
     
  14. Jun 23, 2011 #13
    v = u + 2as, v=0, u=1380m/s ,a= 46

    this formula?
     
  15. Jun 23, 2011 #14

    gneill

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    Nope. Use s = (1/2)at2

    You can either calculate the distance along the straight trajectory and then find the vertical height at the end using the trajectory's angle with the horizontal, or you can calculate the height directly by using in the formula only the vertical component of the acceleration.
     
  16. Jun 23, 2011 #15
    sorry....you are confusing me.....
     
  17. Jun 23, 2011 #16

    gneill

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    Let's see if we can fix that :smile:

    Attached is a diagram of the rocket trajectory. It's a straight line until the fuel runs out after 30 seconds, at which point it will have traveled some distance s along the trajectory and reached height h1 with some speed v. This speed will be directed along the trajectory (angle 70° to the horizontal).

    After burnout the rocket is just a projectile with a ballistic (unpowered) trajectory. It's a parabolic trajectory.

    So there are two sections of the trajectory to worry about. First the powered section, which you are told is a straight line with a given acceleration, and the second is a simple projectile motion with an initial velocity and height and launch angle (70°).

    The first order of business is to find out where the rocket is at the end of the powered section, and how fast it is traveling. Those will in fact be the initial conditions for the second section of the trajectory. The distance it travels in that first section is designated s in the figure. The height it achieves is h1.

    Now, to find the height of the rocket you can either find s using the given acceleration along the straight line for the specified time (30 seconds) and then use trig to find the vertical distance from that, or you can realize that the acceleration has both vertical and horizontal components and use the vertical component of acceleration to find that height. In either case the formula to use is (1/2)at2. In the first case the acceleration is a=46m/s2, directed along the trajectory. In the second case it's the vertical component of a (use the angle and trig to break the 46m/s2 into its vertical and horizontal components).

    Does that help?
     

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    Last edited: Jun 23, 2011
  18. Jun 23, 2011 #17
    now, i really understand the question, you are too great.
    thank you.
    but how to find h2,
    i use this formula, v=u +2as
    u=1380m/s, a=9.8, v=0
     
  19. Jun 23, 2011 #18

    I like Serena

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    Woooooooooooooooowwwww! :smile:
     
  20. Jun 23, 2011 #19
    now, i really understand. you are too great. thank you very much
    but now h2 how to find?
    i use this equation v= u +2as
    v= 0, u= 1380 a=9.8
     
  21. Jun 23, 2011 #20

    gneill

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    First, let's make sure that you've got the first height. What did you get for h1?
     
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