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Homework Help: Projectile motion 2-D

  1. May 28, 2010 #1
    1. The problem statement, all variables and given/known data
    A toy cannon fires a .089kg shell with an initial velocity of 8.9 m/s at an angle of 56* above the horizontal. The shell's trajectory curves downward due to gravity, so at time t=.615s the shell is below the straight line by some vertical distance deltaH. Find this distance deltaH in the absence of air resistance. Answer in units of meters


    2. Relevant equations
    yf=yi+ vyit+1/2ayt2

    xf=xi+ vxit+1/2axt2

    vyi=visin[tex]\vartheta[/tex]

    vxi=vicos[tex]\vartheta[/tex]


    3. The attempt at a solution
    So I have tried two different methods of approaching this problem but still no luck. Here is what I have:

    vyi=8.9m/s(sin56)=7.378434 m/s
    vxi=8.9m/s(cos56)=4.976816 m/s

    [tex]\Delta[/tex]h=7.378434m/s(.615s)-4.9m/s2(.615s)2
    =2.68443 m

    This answer was rejected as incorrect so I tried another method. I figured that I could form a triangle with the given data with vias the hypotenuse and x was found using the following:

    xf=vxit
    =4.976816(.615)=3.06074184 m

    Using this I solved for y in a right triangle as usual with pythagorous at my disposal and obtained a length of 4.53774m which again was rejected.... Any insight as to where I am missing something? Thanks in advance.

    Joe

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 28, 2010 #2

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    I don't think you're interpreting the question correctly.

    This is the vertical distance travelled by the shell. You don't want that. What you want is the difference between the vertical distances that would be travelled with and without the acceleration g. Without g, the shell's trajectory is a straight line. With g, it's a parabola. This question is asking: at time t, what is the vertical separation between the straight line trajectory and the corresponding parabolic one?
     
  4. May 28, 2010 #3
    So then it ought to be 4.9(.615^2)? I guess I can see that since that is how far it would fall after that amount of time correct? Thanks.

    Joe
     
  5. May 29, 2010 #4

    cepheid

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    Science Advisor
    Gold Member

    Yes, because:

    ystraight = vy0t

    and

    yparabolic =vy0t - (1/2)gt2

    Hence, the difference in height between them is given by:

    ystraight - yparabolic = vy0t - [vy0t - (1/2)gt2 ]

    = (1/2)gt2
     
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