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General Launch Angle problem

  1. Feb 16, 2007 #1
    1. The problem statement, all variables and given/known data
    Fairgoers ride a Ferris wheel with a radius of 6.00 m, the bottom of which is 1.75 m from the ground. The wheel completes one revolution every 28.0 s. A passenger drops his keys when he is three quarters of the way up and at an angle of 45° to the vertical. Where do the keys land relative to the base of the ride? (Take the positive direction to be to the right.)


    2. Relevant equations
    x = (Vo cos theta)t
    Vx = Vo cos theta
    Vy = Vo sin theta - gt
    Vox = Vo cos theta
    Voy = Vo sin theta
    Vavg = (2)(Pi)(R/T)


    3. The attempt at a solution
    I first calculated Vo (which, I assume, is constant for a ferris wheel, so Vo = V or Vavg). I really didn't know where to go after that from the above equations, since to calculate x, the distance, we need time. To calculate time, we need Vy. I attempted to calculate initial velocity (Voy = Vo sin theta), which was .95m/s and plug it into the 3rd equation as Vy(as velocity should be constant), to get time... then plug that into the first equation to get the distance... but that didn't seem to work.

    I've tried different formulas and manipulating many things to try and get the answer, but I seem to be stuck. I'm guessing it's something do to with the height and free fall to get t maybe, since I can't seem to get a true t with any of the equations of projectile motion.

    Any assistance would be greatly appreciated. Thanks.
     
  2. jcsd
  3. Feb 16, 2007 #2
    lets see. We have a rotational velocity, which doesn't need to be averaged because its constant. This can be looked at in one of two ways, RPM (or radians/sec where 2*Pi radians= one revolution) or as a point on the circumference moving in two directions at once.

    Point is that the instant the keys are dropped, there are two components to the wheels motion, one upward, and one horizontal. So the keys have initaial velocities in both the x and y directions. The Y component can be treated as throwing up a ball from a given height, how long does it take to hit the ground. From that info, the displacement in the x direction can be calculated as well, but hint: don't forget that it starts left of the base.
    Holler if you need more help.
     
  4. Feb 16, 2007 #3

    berkeman

    User Avatar

    Staff: Mentor

    You are correct to break it into x and y components. The y component will give you the time for the keys to hit the ground, and you use that time in the x component equation (constant horizontal velocity, right?) to tell you how far horizontally the keys move before impact. Just use your 2nd equation for the vertical motion part.

    If you still aren't getting the right answer, post all of your work and we can look to see where you are missing something or have a math error.
     
  5. Feb 16, 2007 #4
    Maybe I'm not thinking about it right. I used the 3rd formula to try and get t, but to get the left side of the equation (Vy), I did 1.34sin45, which is the same as the first part of the right side of the equation, meaning (gt) would be equal to zero... which can't be right. Then tried 1.34 = 1.34sin45-gt and got t= 0.04, which also can't be right since its at least 6m in the air, and g alone wouldn't bring it down that fast. I'm really confused since it seems like Voy should be zero, since the keys aren't being thrown down, just dropped... as for the free fall method, I would guess the height would be 10.75 m, but then it could get into arc length and all that.

    It seems like I'm looking over something very obvious or substituting an incorrect valuable for one of the velocities.
     
  6. Feb 16, 2007 #5

    berkeman

    User Avatar

    Staff: Mentor

    "1.34sin45" is correct for obtaining the initial Vy and Vx (they are equal). So starting with that Vy at the initial height of (what?), how long does it take the keys to travel up (think of it as straight up -- separate the two components in your thinking at first), stop, and then fall back down to the ground. Be careful with your units....stay in mks units throughout.
     
  7. Feb 16, 2007 #6
    Hmm, I'm still getting the wrong answer. The initial height would be 10.75m (I think). Radius + 1/2 Radius + 1.75m = 10.75. The initial Y velocity = .947 m/s. This is where I might be going wrong... I added gravity to the initial Y... so .947 + 9.81 = 10.75 m... so im getting t=1 second.

    Then using x=(1.34cos45)t and getting the initial Y velocity as the distance... .947, which is incorrect.

    The t in seconds would cancel the seconds from the .947 m/s... leaving m which is the correct units.

    Also tried using -.947 since the keys would drop to the left of the center of the ferris wheel, which didn't work.
     
  8. Feb 16, 2007 #7
    The initial heigh which is critical since it starts off the ground, and the keys end up on the ground is 1.75+6(the distance to the hub)+6m(sin 45)= 7.75+.707*6=12.0
     
  9. Feb 16, 2007 #8
    Ok, so with the height being 12m... I got t=1.12s.

    So would the correct answer would be x=(1.34cos45)*1.12 = 1.06?
     
  10. Feb 16, 2007 #9
    Entered 1.06 as my answer and got it wrong. Can someone at least explain it to me so I am able to do a problem like this come time for an exam?
     
  11. Feb 17, 2007 #10
    did you figure in the Y velocity component.

    as in Yf=Y0 + Vy(init)*t + 1/2 at^2

    Vy(init)=1.34*0.707= 0.95
    Yo =12

    quad, see what you get with this. Careful with signs as it will be below Yo
     
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