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Drawing a Y vs Time Graph in a projectile motion problem

  1. Jan 26, 2008 #1
    [SOLVED] Drawing a Y vs Time Graph in a projectile motion problem

    1. The problem statement, all variables and given/known data
    A snowball rolls off a barn roof that slopes downward at an angle of 40 degrees. The edge of the roof is 14.0m above the ground, and the snowball has a speed of 7.00 m/s as it rolls off the roof. Ignore air resistance.

    Draw y-t graphs for the motion in part A.


    2. Relevant equations

    y=(Vo*sin(alpha))*t-1/2*g*t^2

    (In this problem alpha is 40)

    3. The attempt at a solution

    I've tried inserting various time intervals (i.e. .1 sec; .2 sec; .3 sec; etc) to the equation above (I assumed that y= the distance fallen from 14m at the specific time t). So the resultant y I calculated, I subtracted from 14m to get the resultant y coordinate at time t. This being an online assignment, requires exact (y,t) coordinates starting from 0 seconds-1.3 seconds having 0.1 second increments in between. The y coordinates are also specific to 0.1 m increments ranging from 0.0 meters to 14.0 meters. Any possible help would be greatly appreciated. Thanks!
     
    Last edited: Jan 26, 2008
  2. jcsd
  3. Jan 26, 2008 #2
    Bump. Please tell me if I'm doing the right thing and using the correct formula. Or if I have to try a different approach or the formula is incorrectly used.
     
  4. Jan 26, 2008 #3
    y = y0 + v0y*t - (1/2)*g*t^2

    If you consider your referential center at the floor: y0 = 14, the graph is going to decrease from y = 14 to y = 0.
    If you consider your referential center at the edge of the roof: y0 = 0, the graph is going to decrease from y = 0 to y = -14.
    You only need to iterate through t, and find y:)
     
  5. Jan 26, 2008 #4
    yes, and since y is affected by gravity. the graph from y=14 should travel down in a quadratic fashion. So it should look like part of a parabola that opens downward. My only problem is finding specific Y points to plot in the graph.

    If I plug in numbers for t in the equation you used, I should be able to get the correct coords for each y component right? I did this and still the answer wasn't recognized as correct.
     
  6. Jan 26, 2008 #5
    Humm did you notice that v0y is negative in this case right?
     
  7. Jan 26, 2008 #6
    Yes, I used 14-7*sin40*t-1/2*9.8*t^2.

    This still was not the desired answer.
     
    Last edited: Jan 26, 2008
  8. Jan 26, 2008 #7
    This is the resultant graph I had from doing the calculations using the above mentioned formula using a negative velocity. Since we are dealing with trig functions, I made sure that the calculations were done in degrees. It does give a function that somewhat resembles a parabola but it's still not the desired answer. If anyone can help, I'd really appreciate it! :)
     

    Attached Files:

  9. Jan 27, 2008 #8
    Bump! Please can anyone help? If you think this should be the correct answer, tell me that it is so i can at least report that the answer wasn't correctly input for the question.
     
  10. Jan 28, 2008 #9
    hmmm the formula is right, but are you sure that sin function receives the input in degrees?
    Because i used the a sin function that receives the value of the angle in radians and gave it a input in degrees and the graph result is almost the same.
    At t=1,2s gives me 1,544584079m.
     
    Last edited: Jan 28, 2008
  11. Jan 28, 2008 #10
    yeah, it turns out that the online program was bugged and wouldn't accept any answer whatsoever. ha ha. But so long as the equation was right, that's all I care for. The points for the homework aren't as important to me as knowing the actual mechanic of the problem itself. So thanks!
     
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