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Finding an initial speed (projectile motion)

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A rock is thrown from the top of a 20-m high building at an angle of 53 degrees above the horizontal. If the horizontal range of the throw is equal to the height of the building, with what speed was the rock thrown? How long is it in the air?

    Xf = 20, Xi = 0, Vxi = ?, t = ?, Yf = 0, Yi = 20, Vyi = ?, a = -9.81

    2. Relevant equations

    Xf = Xi + Vxi*t
    Yf = Yi + Vyi*t + .5(a)(t)^2


    3. The attempt at a solution

    So first, I tried to find the time. I did this by doing 20 = 0 + Xcos(35)*t where X is initial speed and t is time. I then set t = 20/Xcos(35). I then plugged t back into my Yf equation which gives me 0 = 20 + Xsin(35)*(20/Xcos(35)) - 4.905(20/Xcos(35))^2. Solving this for x gives me about 2.7 m/s which is not right. The answer, according to the answer book, is 11 m/s.


    Can anyone tell me where I'm going wrong?
     
    Last edited: Sep 13, 2009
  2. jcsd
  3. Sep 13, 2009 #2

    rl.bhat

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    0 = Xsin(35)*(20/Xcos(35)) - 4.905(20/Xcos(35))^2.
    In this equation you have not submitted initial y.
     
  4. Sep 13, 2009 #3
    Oops, pardon my typo. I have done it including the Vi in the equation, and 2.7 m/s is what i get. I have changed it above.
     
  5. Sep 13, 2009 #4

    rl.bhat

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    The angle is 53 degrees, not 35 degrees.
    The equation becomes
    0 = 20 + 20*tan53 - 4.9**20^2/vx^2cos^53.
    Now solve.
     
    Last edited: Sep 13, 2009
  6. Sep 13, 2009 #5
    I changed 35 to 53, but it is still incorrect. Is it possible that I am just inputting the formula into my calculator wrong? Maybe this isn't even the way I'm supposed to be solving it? I can't seem to get to 11 m/s no matter how I try.
     
  7. Sep 13, 2009 #6

    rl.bhat

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    I have edited the above post. Now try. You will get the answer.
     
  8. Sep 13, 2009 #7
    That equation can't possibly be correct. You must mean:

    0 = 20 + 20*tan(53)Vxi - 4.9*20^2/Vxi^2*cos(53)^2


    Regardless, this is still incorrect. I must be doing this problem incorrectly. Do you have any suggestions of another method I might take?


    Just for reference, the next part of the problem asks "how long is the rock in the air?" The correct answer in this case is 3.1s, but I do not get this either, obviously.
     
  9. Sep 13, 2009 #8

    rl.bhat

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    [0 = 20 + 20*tan(53)Vxi - 4.9*20^2/Vxi^2*cos(53)^2]
    It is only 20*tan53. There is no vxi there.
     
  10. Sep 13, 2009 #9
    Thank you very much for your patience with me. I finally got the correct answer. It seems the problem was me with my calculator. Thanks again!
     
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