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Projectile motion of a stuntman

  1. Jun 28, 2011 #1
    1. a stuntman performs a stunt of speeding off the cliff of a high rise building with the aim reaching another building with a distance away as the distance between two building is 62m and the differences height between two building is 48m. find the minimum velocity that the car must have as it reached the edge of the roof in order for the stunt to be successful.



    2. first, i find the distance of car from the building to another building = 78.41m
    48^2 + 62^2 = (distance travelled by car)^2
    distance travelled by car= 78.41m
    after that what equation should i use?? and can i assume final velocity = 0m/s?
    help me........thank you


    3. The attempt at a solution
     
  2. jcsd
  3. Jun 28, 2011 #2

    tiny-tim

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    sorry, but this is pointless: the car won't move in a straight line, will it? :redface:

    you need to use the standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations, once in the x direction with acceleration zero, and once in the y direction with acceleration -g …

    what do you get? :smile:
     
    Last edited by a moderator: Apr 26, 2017
  4. Jun 28, 2011 #3
    but the problem is what is the final velocity? zero or other value??
    i know the car falling dowan at 37.75degree from horizontal. i use this formula, tan angle = 48m/62m, hence equal to 37.75degree.
     
  5. Jun 28, 2011 #4

    tiny-tim

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    Last edited by a moderator: Apr 26, 2017
  6. Jun 28, 2011 #5
    you mean should be parabolic path??
    ya... i know...
    v^2=u^2+2as
    v=u +at
    s=ut+at^2
     
  7. Jun 28, 2011 #6

    tiny-tim

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    (try using the X2 tag just above the Reply box :wink:)

    yes :smile:

    (but the last one should be s=ut+ (1/2)at2)

    ok, now do that horizontally, with a = 0, and vertically, with a = -g …

    what do you get? :smile:
     
  8. Jun 28, 2011 #7
    sorry....i know the equation....too panic...
    you mean find the v right now?but i dont know the t.
    and i think a= +g, because toward the ground.
    the correct answer of this question is 19.8m/s.
     
  9. Jun 28, 2011 #8

    tiny-tim

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    you will have two equations (one horizontal and one vertical), and two unknowns (v and t) …

    that should be enough to solve it!

    write out the two equations :smile:
     
  10. Jun 28, 2011 #9
    i tried b4
    s= vt, s=ut+1/2at^2
    t is same
    v is same,
    so,
    s=u(s/v) + 1/2a(s/v)^2
    48m= 62m+ 1/2(9.8) (3844/v^2)
    v=36.68m/s, the correct answer should be19.8m/s
     
  11. Jun 28, 2011 #10

    tiny-tim

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    yes, that's correct :smile:
    no, this is the y equation, so the 48 is correct, but the next term should be uyt, not x :wink:

    (uy is the initial component of velocity in the y direction)

    try again :smile:
     
  12. Jun 28, 2011 #11
    sorry, i do not understand....is the horizontal initial speed is not same as vertical initial speed?
     
  13. Jun 28, 2011 #12

    tiny-tim

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    no no nooo

    the car is initially going at speed v along a flat roof, so the horizontal component is v, and the vertical component is zero

    generally, if the velocity is speed v at an angle θ to the horizontal,

    then the horizontal component of velocity is vcosθ, and the vertical component is vsinθ :smile:

    have you not done components of vectors?
     
  14. Jun 28, 2011 #13
    having dizzy.....
    still cant figure it out....
     
  15. Jun 28, 2011 #14

    tiny-tim

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    uy = 0 (it has no initial vertical speed)

    so uyt = 0

    so … ?​
     
  16. Jun 28, 2011 #15
    ya...s =vt....horizontal moving in the constant velocity. this one i got it.
     
  17. Jun 28, 2011 #16

    tiny-tim

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    yes, s = uxt, which in this case is x = vt, from which you've already got your t = x/v

    but that's the horizontal component equation …

    i'm talking about the vertical component equation, y = uyt + 1/2 gt2 :wink:
     
  18. Jun 28, 2011 #17
    then you mean 48= 0 +1/2(9.8)(62/v cos 37.75)^2
    if like this, i also cant get the answer
     
  19. Jun 28, 2011 #18

    tiny-tim

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    you're losing the plot :redface:

    the plot is, first you find t from the x equation (you got that right, t = x/ux = x/v),

    then you find v from the y equation (using uy = 0)

    the 37.75° is irrelevant
     
  20. Jun 28, 2011 #19
    but i dont know what is v on horizontal line....s=vt, v=?
     
  21. Jun 28, 2011 #20

    tiny-tim

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    i don't understand :confused:

    you've already done this correctly …

    the car is initially moving horizontally, so ux = vcos0° = v, and x = uxt
     
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