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Projectile motion long jumper help

  1. Sep 11, 2009 #1
    1. The problem statement, all variables and given/known data

    An Olympic long jumper is capable of jumping 7.7m. Assuming his horizontal speed is 8.8m/s as he leaves the ground, how long is he in the air? Assume that he lands standing upright--that is, the same way he left the ground.



    2. Relevant equations

    Y = xtan(theta)0 - ((g)/2V02cos2(theta)0)X2

    Vx0 = V0cos(theta)t

    And other that I attempted at using...


    3. The attempt at a solution

    Ok so I have tried several things but, I dont have the angle at which he jumps to figure out the time, and you only get the initial velocity for the horizontal direction. I was gonna use

    Vx0 = V0cos(theta)t

    to solve for Cos(theta) and figure out my angle but I dont even Have V, only Vx so am I suposed to assume they are the same?

    If someone can point me in the right direction, this is the only problem I couldn't figure out earlier today.
     
  2. jcsd
  3. Sep 11, 2009 #2
    Also I cant remeber exactly all the work I did today my work is kind of unorganized and its all over the place but I came to the conclusion that t was 1.11 but that was wrong.
     
  4. Sep 11, 2009 #3

    drizzle

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    Vx0 = V0cos(theta)t

    are you sure of this?, I don't think there's a "t" in that equation
     
  5. Sep 11, 2009 #4
    Oh hold up I mixed up two equation when writing this they were next to eachother on my paper.

    I got Vx0=V0cos(theta)

    and X = V0cos(theta)t

    my bad.
     
  6. Sep 11, 2009 #5
    The only acceleration that the long jumper is subject to is the acceleration due to gravity then no acceleration acts to oppose the long jumpers horizontal motion. Therefore his horizontal velocity remains constant. This problem simply breaks down to d/v=t. 7.7m/(8.8m/s)=.875s
     
  7. Sep 11, 2009 #6
    Wow so simple, I didn't even think about it like that I was trying to do it like the other projectile motion problems I did today. Thank you.
     
  8. Sep 11, 2009 #7
    Alright quick question... Is V the same as Vx0 because there is acceleration in the y direction so there is a vetical velocity component right? Because im trying to now figure out how hight he goes.

    So I figure I use Y= V0sin(theta)t- (1/2)gt2 but I dont have an angle. Which I thoguth I could figure out with X = V0cos(theta)t but I dont know V I only know Vx.
     
  9. Sep 11, 2009 #8

    drizzle

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    no! that'll be true only if his motion was a straight one [as running]. does 7.7 m stands for the height y or the horizontal distance x? if the later, it'll be direct. use the last two equations you’ve stated in your later post to find t, that is:


    Vx0=V0cos(theta)

    and

    X = [V0cos(theta)]t
     
  10. Sep 11, 2009 #9
    7.7m is the distance he jumps in the X direction.
     
  11. Sep 11, 2009 #10

    drizzle

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    great, what's t value?
     
  12. Sep 11, 2009 #11
    The time the way he gave me was correct (we input our answers online, and it accepted it as correct) So I know the time, but how do I get the height without having an angle he jumps at, and I dont know how to get the angle without V since I only have Vx
     
  13. Sep 11, 2009 #12

    drizzle

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    you have x,Vo and t, use the 2nd eq. to get the angle
     
  14. Sep 11, 2009 #13
    The jumper will reach his maximum height at t/2. For a hint. If you need more help let me know.
     
  15. Sep 11, 2009 #14
    You only have V[0][x] not V[0]
     
  16. Sep 11, 2009 #15
    So in this problem is V = Vx? If thats so then

    X = [V0cos(theta)]t

    Then the I would devide by t and V0 right? which gives me .9943 = cos(theta) so then inverse cos of .9943 would give me my angle? which I got to be 6.11 which doesent seem right, but I then used that and the Y I got was incorrect.

    ok just saw the post that says you got Vx not V and then how do I find the angle without V?
     
  17. Sep 11, 2009 #16
    I have two attempts left on the problem and I dont know how to find the angle. So I can't guess anymore, need some hints. I mean I know if you want to maximize your X you would jump at a angle of 45 but not sure if you can assume that here and dont want to waste my attempt trying it.
     
  18. Sep 11, 2009 #17
    Ok on my last attempt, someone please help out before I screw it up. How can you find the angle if you dont know V0?
     
  19. Sep 11, 2009 #18
    Damn used up my last attempt and wasn't able to get the maximum height.
     
  20. Sep 11, 2009 #19

    drizzle

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    type the eq. of Vy!
     
  21. Sep 11, 2009 #20

    rl.bhat

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    The projectile equation is
    Y = xtan(θ) - (g)x^2/2(V0^2cos^2(θ)
    Τhe net displacement is zero. Hence
    xtan(θ) = (g)x^2/2(V0^2cos^2(θ) or
    tan(θ) - (g)x/2(V0^2cos^2(θ) or
    tan(θ) = gx/2(vx)^2
    Substitute the values and find θ.
     
    Last edited: Sep 11, 2009
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