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Projectile Motion

  1. Apr 2, 2007 #1
    1. The problem statement, all variables and given/known data
    An object was launched from the top of a cliff at some angle with an acceleration of 50 ft/sec that overcame gravity. The fuel tank ran out after 2 seconds of flight and gravity took over from that point forward. The object reached a maximum height of 310 ft, was in the air for a total of 9.16 seconds and traveled a horizontal distance of 383 ft.
    Initial Launch Angle
    Height of the cliff
    Initial vertical velocity
    Initial horizontal velocity

    2. Relevant equations


    3. The attempt at a solution
    I think I may have found the launch angle by using the 1/2at^2 equation and plugging in 383=1/2ax(9.16)^2 and 310=1/2ay(9.16)^2. I found the angle to be 39.36, but I am unsure if it is correct, after that I am stuck, help is greatly appreciated.
  2. jcsd
  3. Apr 2, 2007 #2
  4. Apr 2, 2007 #3
    lets see, the first thing to sieze on is the horizontal displacement:

    if its 383' and t=9.16 we know, Vx=41.8'/s (ave velocity)

    moreover we know that the force/m applied was 5*g, and the duration 2 seconds.

    The initial x velocity of rocket 0.

    so for the x component we have a displacement of two components:

    that covered during acceleration phase and that while coasting here a is the x component of acceleration

    383=(1/2at^2)+ at(t') where t+t'=9.16 and since t=2, t'=7.16, hence a

    383=.5a*(2^2)+a*2(7.16) we have 2a+14.32a=383 or 16.32a=383 a=23.47

    so atotal/ax=5*32.2/23.47=6.86 ie arccos=1/6.86=81 degrees. Sure you can manage from here. This any help?
    Last edited: Apr 2, 2007
  5. Apr 2, 2007 #4
    thanks, having the proper angle is a big help...im having a tough time with the height though...
  6. Apr 2, 2007 #5
    If its launched at an angle theta, then for the first two seconds, its also accelerating at 50 ft/sec^2. So, resolving the acceleration into horizontal and perpendicular components, you can find the initial speed in the x (horizontal) direction.

    Now, in the vertical direction, there are two forces/accelerations acting. Their vector sum will give you the net acceleration, and this (over a period of 2 seconds) will give you the initial speed in the y direction.

    Since you know what the max height is, you can use the above to get an equation in theta/time. And using the horizontal distance you get another equation in theta/time. Two equations and two variables, you get your answer.
  7. Apr 2, 2007 #6
    sorry, but I don't follow.....I have it set up now as 383=1/2ax(9.16)^2 and 310=1/2ay(9.16)^2, is that what you are saying?
  8. Apr 2, 2007 #7
    well its a bear of a problem. ugliest w/o air resistance I've seen in a while. But I can't do all the work. So a hint will have to suffice:
    parce it out for the two phases, with ay =sin81(5*g) at 2 seconds for phase 1, then it becomes purely kinematic. so you need to keep track of the altitude above the cliff at 2 seconds, and the fact during purely ballistic phase it starts at that altitude and lands below both the launch site and the time where it became purely ballistic.
  9. Apr 2, 2007 #8
    thanks, im going to work on it for a little while longer and then call it a night...lol
  10. Apr 2, 2007 #9
    have fun, its pugugly but doable. You'll be a manlier man for having done so?? dont forget gravity during both phases!!!!
  11. Apr 3, 2007 #10


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    Staff Emeritus
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    Exactly HOW did you find the launch angle? Please show your work.
    The maximum height is reached in 9.16 seconds but NOT its full horizontal distance!
  12. Apr 3, 2007 #11
    any more help with this? I just keep getting stuck...
  13. Apr 3, 2007 #12
    please note mistake above from english units--I put in 5*g for acceleration when it was 50'/sec^2. My bad. So much more used to Mks. The approach is the same. the angle should be 50/23.6=1/arccos=61.8 degrees, sorry.
  14. Apr 3, 2007 #13


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    Homework Helper

    Also, if you haven't done it already, it REALLY helps to draw a picture labeling all relevant information. It helps relate the equations to the physical situation and keep signs straight. Free Body Diagram........
  15. Apr 3, 2007 #14
    I finally got the problem finished earlier today, I will post my results up later...it was pretty tough...thanks for help
  16. Apr 3, 2007 #15
    yea, I found that to be the angle also, I finished up the problem earlier, I will post up my results soon, thanks for help...
  17. Apr 3, 2007 #16


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    Staff Emeritus
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    Gold Member

    I've merged your two threads.
  18. Apr 3, 2007 #17
    thanks, i looked at this tonite with all the interdigitatated posts and wondered whether it was me or the Twilight zone that PF had entered.
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