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(Projectile motion) Little question on a problem I posted 5 days ago

  1. Nov 23, 2005 #1
    I posted the following problem a few days ago: So it's about somebody on a bike who rides off an entrenchment (that's what it's called right?) with a velocity v under an angle of alpha with the ground. He's hoping to land safely on another entrenchement that's h heigher than the first one, at a distance x from the first entrenchment:
    [​IMG]
    For a given height h, I found the minimal velocity vmin the jumper needs to have in order to land safely on the platform at a distance x:
    v0= (x/cos@) sqrt(g/2(x tan @ -h))

    But how can I show that no matter what his takeoff speed is, the maximum height of the platform is hmax < x tan @. I need to interpret this result physically. Well I don't know how to do the last thing, because I know that is hmax < x tan @ the argument under the sqrt will be negative and therefore out of the domain of the sqrt function... I don't see the physical consequences... Could someone please help me?
     
  2. jcsd
  3. Nov 23, 2005 #2

    Tide

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    This is just a geometry problem. If you project a straight line along the lower ramp then that defines the greatest hmax.
     
  4. Nov 23, 2005 #3
    Could you explain it a little bit more please ... I still don't quite get it.... :( sorry
     
  5. Nov 23, 2005 #4

    Tide

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    Draw a line extending the left ramp in your diagram ... now imagine the right ramp being taller and taller. Eventually, its top corner will be above the line you created. No matter how fast the projectile leaves the left ramp, it will not be able to land safely on the right side!
     
  6. Nov 23, 2005 #5
    Ok but I need to explain with the formula v0= (x/cos@) sqrt(g/2(x tan @ -h)) why the maximum height of the platform needs to be hmax < x tan @, no matter what the v0 is. This result needs to be interpreted physically... How would I do that?
     
  7. Nov 23, 2005 #6
    All I know is that if hmax > x tan @ than the argument under the sqrt will be negative and therefore out of the domain of the sqrt function... I don't see the physical consequences...
     
  8. Nov 23, 2005 #7
    Well, this is correct because it proves that you cannot calculate v_0 when hmax>xtan@

    Besides, just look at the triangle you get when i draw a straight line from the left to the right entrenchment, in the direction of v_0. Let us call that distance R.

    You get that
    1) h = Rsin@
    2) x = Rcos@

    Devide 1) by 2) and you are done

    marlon

    ps : vanwaar in Nederland zijt gij en waar studeert ge ?
     
  9. Nov 23, 2005 #8
    Yeah but what are the physical consequences? (I don't know what they want to hear when they ask for the physical consequences)

    ps: Amsterdam
     
  10. Nov 23, 2005 #9
    Physical consequences just means that you cannot calculate the velocity in the case of hmax > xtan@. If you cannot calculate the velocity, there is NO MOTION possible because you get an imaginary number. That is all

    Ahh, lang geleden dat ik daar nog geweest ben. Ikzelf woon in Gent , België.

    Bedankt voor het antwoord

    regards
    marlon
     
  11. Nov 23, 2005 #10
    Ohww dat t zo simpel is :P Ik zat er helemaal stuff achter te zoeken. Bedankt in ieder geval voor je antwoord! Succes met je studie!
     
  12. Nov 23, 2005 #11
    Als je nog studeert dan he ;)
     
  13. Nov 23, 2005 #12
    ik doctoreer...mmm da is nie echt meer studeren :wink:

    marlon
     
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