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Calculate the clown's velocity as he lands in the net

  1. Feb 26, 2007 #1
    1. The problem statement, all variables and given/known data

    A circus clown is fired from a cannon into a net that is situated 2.0 m above the cannon at some distance from it. The cannon is elevated at 50 degrees to the horionztal and the clown's speed at lunch if 15 m/s.
    (a) Find the horizontal distance from the cannon where the net needs to be placed in order for the clown to land in it.
    (b) Calculate the clown's velocity as he lands in the net


    2. Relevant equations

    a = v/t
    v2 = v1 + at
    d = (v1 + v2 / 2) * t
    d = v1t + 1/2at^2
    d = v2t - 1/2at^2
    v2^2 = v1^2 + 2ad

    3. The attempt at a solution

    (a) To calculate the horizontal velcoity you do 15cos50 = 9.642 m/s
    for the vertical velocity you do 15sin50 = 11.491 m/s

    dv = vv1t + 1/2gt^2
    -2 = 11.491t - 4.9t^2
    4.9t^2 - 11.491t - 2 = 0
    by using the quadratic formula you get
    t = 6.345 s

    subing the time into d = vt
    d = 9.642(6.345)
    d = 61.178 metres

    (b) vv2^2 = vv1^2 + 2gdv
    vv2^2 = 11.491^2 + 2(-9.8)(2)
    vv2 = -9.6

    v2^2 = vv1^2 + vv2^2
    v2 = 15.3

    tan theta = 9.6 / 9.642
    theta = 45

    the clowns velocity has he hits the net is 15.3 m/s [36 degrees E of S]

    I am pretty sure I did part a right but I am not sure about part b.
     
  2. jcsd
  3. Feb 27, 2007 #2

    hage567

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    Homework Helper

    (a) I'm not sure I understand the h = -2 m. The net is 2 m above the ground. I think you might be confusing the negative acceleration due to gravity with direction. :confused: Other than that, your method is correct.
    (b) If you look at the symmetry of the trajectory what do you notice about your answer that might indicate it is incorrect? If the clown intercepts the net at a height above the ground, his velocity should be less than what he started out at. Does that make sense? To solve (b), consider the problem from the time he reaches maximum height in his flight, to the net. What can you say about the velocity at the maximum height?
     
  4. Feb 27, 2007 #3
    Yeah, the distance should be 2 m, not -2 m for part a.

    I am not sure if I am following you for part b though. I understand how his velocity should be less than when he started. His vertical velcoity at maximum height would be 0 and his horizontal velocity would still be the same, which is 9.6m/s.

    Here is my second attempt:

    (a) To calculate the horizontal velcoity you do 15cos50 = 9.642 m/s
    for the vertical velocity you do 15sin50 = 11.491 m/s

    dv = vv1t + 1/2gt^2
    2 = 11.491t - 4.9t^2
    4.9t^2 - 11.491t + 2 = 0
    by using the quadratic formula you get
    t = 2.1 or 0.19, We would use the 2.1 since the 0.19 would be the 2m right after take off.

    subing the time into d = vt
    d = 9.642(2.1)
    d = 20.25 metres

    (b) vv2^2 = vv1^2 + 2ad
    vv2^2 = 11.491^2 + 2(-9.8)(2)
    vv2 = +- 9.637

    v2^2 = vh^2 + vv2^2
    v2 = 13.6

    tan theta = 9.642 / 9.637
    theta = 45

    The second part is still confusing me :S
     
    Last edited: Feb 27, 2007
  5. Feb 27, 2007 #4

    hage567

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    Homework Helper

    What you need is to find the time that the clown falls after reaching max height (when Vy=0) to when he hits the net. You already know how long it takes him to get to the net since you found that in part (a). The difference of these times is how long he fell after reaching max height (when Vy=0), to the net. Once you have this time, you can use it to find his velocity after falling from max height to the net.
    How do you think you can find the time to maximum height? Think about the symmetry of parabolic motion again. Also, I wouldn't use the equation "vv2^2 = vv1^2 + 2ad" in your calculations, it isn't really suitable in this case.
    See if that helps you out.
     
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